Alhamdulillah …. Marhaban Ya Ramadhan

Alhamdulillah kita masih dipertemukan dengan bulan yang penuh maghfiroh ini yakni bulan Ramadhan 1438 H semoga kita dapat mengisinya dengan berbagai amal solih dan semoga kita juga diberikan kesempatan untuk dapat memperbanyak amal solih tersebut. Dan juga semoga kita diberikan kekuatan oleh Allah untuk  dapat terhindar dari perbuatan-perbuatan maksiat. amiin

Marhaban ya Ramadhon

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Contoh Soal Turunan Fungsi

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Penggunaan Turunan Fungsi (Lanjutan Turunan Fungsi)

Terdapat pada :

  • Persamaan garis singgung
  • Fungsi naik dan fungsi turun
  • menggambar grafik fungsi aljabar
  • Maksimum dan minimum fungsi
  • Teorema L’Hopital  (dibaca: Lopital)
  • Nilai stasioner
  • Titik belok
  • Kecepatan dan percepatan

Perhatikanlah tabel berikut!

\begin{array}{|c|c|l|c|}\hline \multicolumn{3}{|c|}{\textrm{Turunan Pertama}}&\textrm{Turunan Kedua}\\\hline 1.&\textrm{\textrm{Gradien garis singgung}}&m={f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}=\displaystyle \frac{f(x+h)-f(x)}{h}&\textrm{Belok}\\\cline{1-3} 2.&\textrm{Fungsi naik dan turun}&y=f(x)\begin{cases} {f}'(x)> 0, & \text{ fungsi naik } \\ {f}'(x)< 0, & \text{ fungsi turun } \end{cases}&\textrm{Percepatan}\\\cline{1-3} 3.&\textrm{Jarak, kec, percepatan}&y=s(t)\begin{cases} s(t) & \text{ jarak} \\ {s}\, '(t) & \text{ kecepatan } \\ {s}\, ''(t) & \text{ percepatan} \end{cases}&\textrm{Maksimum}\\\cline{1-3} &&\begin{aligned}\textrm{Maksimum}:&\\ \rightarrow {f}&\, ''(k)< 0\\ \textrm{titik mak}&\: \left ( k, f(k)\right ) \end{aligned}&\textrm{Minimum}\\\cline{3-3} 4.&\textrm{Stasioner}&\begin{aligned}\textrm{Minimum}:&\\ \rightarrow {f}&\, ''(k) > 0\\ \textrm{titik min}&\: \left ( k, f(k)\right ) \end{aligned}&\\\cline{3-3} &{f}\, '(x)=0\rightarrow x=k&\begin{aligned}\textrm{Belok}:&\\ \rightarrow {f}&\, ''(k)= 0\\ \textrm{titik belok}&\: \left ( k, f(k)\right ) \end{aligned}&\\\cline{1-3} 5.&\begin{aligned}&\textrm{Limit fungsi}\\ &\textrm{bentuk tak tentu} \end{aligned}&\begin{aligned}&\textrm{Aturan L'Hopital}\\ &\\ &\underset{x\rightarrow h}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)}{g(x)}=\underset{x\rightarrow h}{\textrm{Lim}}\: \: \displaystyle \frac{{f}\, '(x)}{g\,'(x)}\\ &\\ &\textrm{untuk hasil limit}\\ &\textrm{bentuk}\: \: \frac{0}{0}\: \: \textrm{atau}\: \: \frac{\infty }{\infty }\end{aligned}&\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Tentukanlah persamaan garis singgung pada kurva}\: \: y=x^{2}+2x-8\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: 1 \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|l|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Diketahui},\: &\textrm{persamaan sebuah kurva adalah:}\\ &y=x^{2}+2x-8\\ & \end{aligned}}\\\hline \textrm{Titik singgung}&\textrm{Gradien}\: ,\: x=1&\textrm{Persamaan garis singgung}\\\hline \begin{aligned}x=1\rightarrow y&=(1)^{2}+2(1)-8\\ &=1+2-8\\ &=-5\\ &\\ \textrm{di titik}&\: \: (a,b)=(1,-5)\end{aligned}&\begin{aligned}\displaystyle \frac{dy}{dx}=m&=2x+2\\ &=2(1)+2\\ &=4\\ &\\ & \end{aligned}&\begin{aligned}y&=m(x-a)+b\\ &=4(x-1)+(-5)\\ &=4x-4-5\\ &=4x-9\\ & \end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Jadi},\: &\textrm{persamaan garis singgungnya adalah:}\\ &y=4x-9\Leftrightarrow y-4x+9=0\\ & \end{aligned}}\\\hline \end{array}.

Berikut ilustasi gambar dari persoalan di atas

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah interval di mana kurva fungsi}\: \: f(x)=x^{3}+3x^{2}-9x+5\\ &\textrm{a}.\quad \textrm{naik}\\ &\textrm{b}.\quad \textrm{turun} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ f(x)&=x^{3}+3x^{2}-9x+5\\ {f}\, '(x)&=3x^{2}+6x-9\\ &=3(x+3)(x-1)\\ & \end{aligned}}\\\hline \textrm{naik}\: ;\: \: {f}\, '(x)> 0&\textrm{turun}\: ;\: \: {f}\, '(x)< 0\\\hline \begin{aligned}&\\ &3(x+3)(x-1)> 0\\ & \end{aligned}&\begin{aligned}&\\ &3(x+3)(x-1) < 0\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{array}{ll|lll|lll}\\ &\multicolumn{3}{c}{.}&\multicolumn{3}{c}{.}&\\\cline{1-2}\cline{6-7} +&+&-&-&-&+&+&\\ &&&&&&\\\hline &\multicolumn{2}{l}{-3}&&\multicolumn{2}{c}{1}&&\\ \end{array} }\\\hline \begin{aligned}&\\ \textrm{naik}\: ,&\: x< -3\: \: \textrm{atau}\: \: x> 1\\ & \end{aligned}&\begin{aligned}&\\ \textrm{turun}\: ,&\: -3< x< 1\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah nilai stasioner fungsi}\: \: f(x)=x^{3}+3x^{2}-9x+5 \: \: \textrm{dan tentukan pula jenisnya}.\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}&\\ f(x)&=x^{3}+3x^{2}-9x+5\\ {f}\, '(x)&=3x^{2}+6x-9\\ &=3(x+3)(x-1)\\ &\textrm{dan}\\ &\begin{array}{ll|lll|lll}\\ &\multicolumn{3}{c}{.}&\multicolumn{3}{c}{.}&\\\cline{1-2}\cline{6-7} +&+&-&-&-&+&+&\\ &&&&&&\\\hline &\multicolumn{2}{l}{-3}&&\multicolumn{2}{c}{1}&&\\ \end{array} \\ &\textrm{untuk nilai stasionernya} \\ f(-3)&=(-3)^{3}+3(-3)^{2}-9(-3)+5=32&\rightarrow \left ( -3,32 \right )\: \textrm{adalah titik balik maksimum}\\ f(1)&=(1)^{3}+3(1)^{2}-9(1)+5=0&\rightarrow \left ( 1,0 \right )\: \textrm{adalah titik balik minimum} \end{aligned}.

Berikut ilustrasi gambar kurvanya baik untuk jawaban No. 2 maupun No. 3

\begin{array}{ll}\\ 4.&\textrm{Masih sama dengan soal seperti pada No. sebelumnya, yaitu fungsi}\: \: f(x)=x^3+3x^2-9x+5\: .\\ & \textrm{Tentukanlah koordinat titik beloknya}\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|c|c|l|c|}\hline \multicolumn{6}{|c|}{\begin{aligned}&\\ f(x)&=x^3+3x^2-9x+5\\ {f}\, '(x)&=3x^{2}+6x-9\\ {f}\, ''(x)&=6x+6\\ \textrm{Proses}\: &\textrm{mencari beloknya}\\ {f}\, ''(x)&=0\\ 6x+6&=0\\ 6x&=-6\\ x&=-1\\ & \end{aligned}}\\\hline \textrm{Interval}&f(x)&{f}\, '(x)&{f}\, ''(x)&\: \: \qquad \textrm{Keterangan}&\textrm{Koordinat titik beloknya}\\\hline x< -1&&&-&\textrm{grafik cekung ke bawah}&\\\cline{1-5} x= -1&16&-12&0&\textrm{grafik memiliki titik belok}&(-1,16)\\\cline{1-5} x > -1&&&+&\textrm{grafik cekung ke atas}&\\\hline \end{array}.

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Turunan Fungsi (KTSP MA/SMA Kelas XI)


A. Turunan Fungsi

A. 1  Pendahuluan

mengenal laju perubahan untuk nilai fungsi

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Laju Perubahan}}\\\hline \textrm{Laju erubahan rata-rata}&\textrm{Laju perubahan sesaat}\\\hline \begin{aligned}&\\ &\displaystyle \frac{\Delta y}{\Delta x}=\displaystyle \frac{f\left ( x_{2} \right )-f\left ( x_{1} \right )}{x_{2}-x_{1}}\\ & \end{aligned}&\begin{aligned}&\\ &\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(a+h)-f(a)}{h}\\ & \end{aligned}\\\hline \end{array}.

A. 2  Pengertian

Perhatikanlah ilustrasi berikut!

Misalkan diketahui fungsi   y=f(x)   terdefinisi untuk semua harga x di sekitar  x=k . Jika  \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}  ada, maka bentuk  \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}  disebut turunan dari fungsi  f(x)  saat  x=k.

A. 3  Notasi

  • Notasi turunan fungsi dilambangkan dengan  {f}'(k)   dengan  {f}'(k)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}.
  • Lambang  {f}'(k)  dibaca  f  aksen  k  desebut  turunan atau derivatif  untuk fungsi  f(x)   terhadap  x   saat   x=k.
  • Jika limitnya ada, dapat dikatakan fungsi  f(x)   diferensiabel(dapat didiferebsialkan)  saat  x=k   dan bentuk limitnya selanjutna dilambangkan dengan  {f}'(k).
  • Misalkan fungsi  f(x)  memiliki turunan  {f}'(x) . Jika   {f}'(k)   tidak terdefinisi  maka  f(x)   tidak diferensiabel  di   x=k .

A. 4  Bentuk Umum (Turunan Pertama)

Bentuk umum turunan fungsi ang selanjutnya disebut juga turunan pertama  fungsi  y   terhadap  x  dapat dinotasikan dengan berbagai bentuk berikut yaitu:

{y}\: '=\displaystyle \frac{dy}{dx}=\displaystyle \frac{d\left ( f(x) \right )}{dx}={f}'(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Jika} \: \: g(x)=3x-5, \textrm{hitunglah laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{g(x)=3x-5}\\\hline \textrm{Cara Pertama}&\textrm{Cara Kedua}\\\hline \begin{aligned} g(x)&=3x-5\\ g(2)&=3.2-5=1\\ {g}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{g(x)-g(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(3x-5)-(1)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{3x-6}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 3\\ &=3 \end{aligned}&\begin{aligned}g(2)\: \,&=1\\ g(2+&h)=3(2+h)-5=3h+1\\ g(2+&h)-g(2)=3h\\ {g}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{g(2+h)-g(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{3h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 3\\ &=3\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{Jadi, laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\: \: \textrm{adalah 3}}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: f(x)=2017x^{2},\: \: \textrm{tentukanlah}\: \: {f}'(x)\: \: \textrm{dan}\: \: {f}'(1)\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline {f}'(x)&{f}'(1)\\\hline \begin{aligned}f(x)&=2017x^{2}\\ f(x+h)&=2017(x+h)^{2}\\ &=2017\left ( x^{2}+2xh+h^{2} \right )\\ &=2017x^{2}+4034xh+2017h^{2}\\ {f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (2017x^{2}+4034xh+2017h^{2} \right )-\left (2017x^{2} \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4034xh+2017h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 4034x+2017h\\ &=4034x \end{aligned}&\begin{aligned}{f}'(x)&=4034x\\ \textrm{maka},&\\ {f}'(1)&=4034.1\\ &=4034\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\end{array}.

\begin{array}{ll}\\ 3.&\textrm{Diketahui bahwa fungsi}\: \: f(x)=\displaystyle \frac{1}{x^{2}}\: \: \textrm{dengan daerah asal}\: \: \textrm{D}_{f}=\left \{ x\: |\: x\in \mathbb{R},\: \: \textrm{dan}\: \: x\neq 0 \right \}.\\ &\textrm{a}.\quad \textrm{Tunjukkan bahwa}\: \: {f}'(a)=\displaystyle -\frac{2}{a^{3}}\\ &\textrm{b}.\quad \textrm{jelaskanlah mengapa}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|l|l|}\hline \begin{aligned}{f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{(x+h)^{2}}-\displaystyle \frac{1}{x^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}-(x+h)^{2}}{\left (x(x+h) \right )^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-\left ( x^{2}+2xh+h^{2} \right )}{h\left ( x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2xh-h^{2}}{h\left (x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: -\displaystyle \frac{2x+h}{\left (x(x+h) \right )^{2}}\\ &=-\displaystyle \frac{2x}{x^{4}}\\ &=-\displaystyle \frac{2}{x^{3}} \end{aligned}&\begin{aligned}\textrm{Untuk}&\: \textrm{jawaban poin a dan b adalah sebagai berikut}\\ {f}'(x)&=-\displaystyle \frac{2}{x^{3}}\\ {f}'(a)&=-\displaystyle \frac{2}{a^{3}}\\ \textrm{maka},&\\ {f}'(0)&=-\displaystyle \frac{2}{0^{3}}\\ &=-\displaystyle \frac{2}{0}\\ &\\ &\textrm{karena penyebut berupa bilangan}\: \: 0\\ &\textrm{maka}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Selanjutnya berikut beberapa hasil turunan dari berbagai fungsi aljabar}\\ &\\ &\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|}\hline y&\cdots &x^{-2}&x^{-1}&C&x&x^{2}&x^{3}&x^{4}&x^{5}&\cdots \\\hline \multicolumn{11}{|c|}{\begin{aligned}&\\ &\textrm{dan hasilnya}\\ & \end{aligned}}\\\hline {y}\: '&\cdots &-\displaystyle \frac{2}{x^{3}}&-\displaystyle \frac{1}{x^{2}}&0&1&2x&3x^{2}&4x^{3}&5x^{4}&\cdots \\\hline \end{array} \end{array}.

B. Rumus Turunan

\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Untuk}:&\\ &\begin{cases} a & \in \mathbb{R} \\ n& \in \mathbb{Q}\\ c& \textrm{konstanta}\\ U&=g(x)\\ V&=h(x) \end{cases}\\ & \end{aligned}}\\\hline \textrm{Sifat-Sifat}&\textrm{Fungsi Aljabar}&\textrm{Fungsi Trigonometri}\\\hline \begin{aligned}y&=c\rightarrow &{y}'&=0\\ y&=c.U\rightarrow &{y}'&=c.{U}'\\ y&=U\pm V\rightarrow &{y}'&={U}'\pm {V}' \end{aligned}&\begin{aligned}y&=a.x^{n}\rightarrow {y}\: '=n.a.x^{(n-1)}\end{aligned}&\begin{aligned}y&=a\sin x\rightarrow & {y}'&=a\cos x\\ y&=a\cos x\rightarrow &{y}'&=-a\sin x\\ y&=a\tan x\rightarrow &{y}'&=a\sec ^{2}x\end{aligned}\\\cline{2-3} \begin{aligned}y&=U.V\rightarrow &{y}'&={U}'.V+U.{V}'\\ y&=\displaystyle \frac{U}{V}\rightarrow &{y}'&=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}&\begin{aligned}y&=a.U^{n}\rightarrow {y}\: '=n.a.U^{(n-1)}.{U}'\end{aligned}&\begin{aligned}y&=a\sin U\rightarrow & {y}'&=\left (a\cos U \right ).{U}'\\ y&=a\cos U\rightarrow &{y}'&=\left (-a\sin U \right ).{U}'\\ y&=a\tan U\rightarrow &{y}'&=\left (a\sec ^{2}U \right ).{U}'\end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Aturan}&\: \: \textrm{rantai untuk turunan pada} \: \: y=f(u),\: \: \textrm{maka}\\ \textrm{untuk}\: \: &u\: \: \textrm{merupakan fungsi}\: \: x,\: \: \textrm{adalah}:\\ &\\ {y}\: '&={f}\: '(x).{u}'\\ &\textrm{atau}\\ \displaystyle \frac{dy}{dx}&=\displaystyle \frac{dy}{du}.\frac{du}{dx}\\ & \end{aligned}}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Dengan menggunakan rumus turunan}\: \: {f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h},\: \: \textrm{tunjukkan bahwa turunan}\: \: \\ &\textrm{a}.\quad f(x)=ax^{n}\: \: \textrm{adalah}\: \: {f}\, '(x)=n.ax^{n-1}\\ &\textrm{b}.\quad f(x)=u(x)+v(x)\: \: \textrm{adalah}\: \: {f}\, '(x)= {u}\,'(x)+{v}\,'(x)\\ &\textrm{c}.\quad f(x)=u(x).v(x)\: \: \textrm{adalah}\: \: {f}\, '(x)= {u}\,'(x).v(x)+u(x).{v}\,'(x)\\ &\textrm{d}.\quad f(x)=\displaystyle \frac{u(x)}{v(x)}\: \: \textrm{adalah}\: \: {f}\, '(x)=\displaystyle \frac{{u}\,'(x).v(x)+u(x).{v}\,'(x)}{(v(x))^{2}} \\ &\textrm{e}.\quad f(x)=\sin x\: \: \textrm{adalah}\: \: {f}\, '(x)=\cos x \\ &\textrm{f}.\quad f(x)=\cos x\: \: \textrm{adalah}\: \: {f}\, '(x)=-\sin x \\ &\textrm{g}.\quad f(x)=\tan x\: \: \textrm{adalah}\: \: {f}\, '(x)=\sec^{2} x \\ &\textrm{h}.\quad f(x)=\cot x\: \: \textrm{adalah}\: \: {f}\, '(x)=-\csc^{2} x \\ \end{array}.

Bukti:

\begin{array}{ll}\\ 1.\: \textrm{a}&{f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\, '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a(x+h)^{n}-ax^{n}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left ( x^{n}+\binom{n}{1}x^{n-1}.h+\binom{n}{2}x^{n-2}.h^{2}+\binom{n}{3}x^{n-3}.h^{3}+\cdots +\binom{n}{n-1}x.h^{n-1}+h^{n}\right )-ax^{n}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left (\binom{n}{1}x^{n-1}.h+\binom{n}{2}x^{n-2}.h^{2}+\binom{n}{3}x^{n-3}.h^{3}+\cdots +\binom{n}{n-1}x.h^{n-1}+h^{n}\right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\left (\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}.h+\binom{n}{3}x^{n-3}.h^{2}+\cdots +\binom{n}{n-1}x.h^{n-2}+h^{n-1}\right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle a\binom{n}{1}x^{n-1}+a\binom{n}{2}x^{n-2}.h+a\binom{n}{3}x^{n-3}.h^{2}+\cdots +ah^{n-1}\\ &=a\binom{n}{1}x^{n-1}+0+0+...+0\\ &=a.\displaystyle \frac{n!}{(n-1)!.1!}x^{n-1}\\ &=a.\displaystyle \frac{n.(n-1)!}{(n-1)!}x^{n-1}\\ &=a.n.x^{n-1}\qquad\quad \blacksquare \end{aligned} \end{array}.

\begin{array}{ll}\\ 1.\: \textrm{g}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\tan x+\tan h}{1-\tan x.\tan h}-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\tan x+\tan h-\tan x+\tan^{2}x.\tan h}{1-\tan x.\tan h}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan h\left ( 1+\tan^{2}x \right )}{h\left ( 1-\tan x.\tan h \right )}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan h}{h}\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{1+\tan^{2}x}{1-\tan x.\tan h}\\ &=1 \times \displaystyle \frac{1+\tan^{2}x}{1-0}\\ &=1+\tan^{2}x\\ &=\sec ^{2}x\qquad\quad \blacksquare \end{aligned} \end{array}.

Untuk bukti yang lain silahkan dicoba sebagai latihan mandiri.

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah turunannya}\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=x^{6}&\textrm{i}.&f(x)=(x+1)(x-2)&\textrm{q}.&f(x)=\displaystyle \frac{2}{x^{3}}\\ \textrm{b}.&f(x)=\displaystyle \frac{1}{2}x^{2}&\textrm{j}.&f(x)=(3-x)(5-x)&\textrm{r}.&f(x)=\displaystyle \frac{1}{2\sqrt{x}}\\ \textrm{c}.&f(x)=-2x^{5}&\textrm{k}.&f(x)=(2x+3)^{2}&\textrm{s}.&f(x)=\displaystyle \frac{1}{3x^{5}}\\ \textrm{d}.&f(x)=ax^{3}&\textrm{l}.&f(x)=(x-2)^{3}&\textrm{t}.&f(x)=2x^{4}+\displaystyle \frac{1}{2x^{3}}\\ \textrm{e}.&f(x)=4x^{4}-x^{2}+2017&\textrm{m}.&f(x)=(4x+1)(4x-1)&\textrm{u}.&f(x)=\displaystyle \frac{x}{2}+\frac{2}{x}\\ \textrm{f}.&f(x)=6-x-3x^{2}&\textrm{n}.&f(x)=(x-1)(x+1)(x+2)&\textrm{v}.&f(x)=\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )^{2}\\ \textrm{g}.&f(x)=(x-3)^{2}&\textrm{o}.&f(x)=\displaystyle \frac{1}{2}x^{-4}&\textrm{w}.&f(x)=x^{2}\left ( 1+\sqrt{x} \right )^{2}\\ \textrm{h}.&f(x)=\left ( x^{3}-2 \right )^{2}&\textrm{p}.&f(x)=x^{-5}&\textrm{x}.&f(x)=\displaystyle \frac{2+4x}{x^{2}}\\ &&&&\textrm{y}.&f(x)=\left ( x+1+\displaystyle \frac{1}{x} \right )\left ( x+1-\displaystyle \frac{1}{x} \right ) \end{array} \end{array}.

Jawab:

Untuk  y=f(x)  maka ,

\begin{array}{|c|c|c|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}&\textrm{d}&\textrm{e}&\textrm{f}\\\hline \begin{aligned}y&=x^{6}\\ {y}\: '&=6x^{6-1}\\ &=6x^{5} \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{2}x^{2}\\ {y}\: '&=2.\displaystyle \frac{1}{2}x^{2-1}\\ &=x \end{aligned}&\begin{aligned}y&=-2x^{5}\\ {y}\: '&=-5.2x^{5-1}\\ &=-10x^{4} \end{aligned}&\begin{aligned}y&=ax^{3}\\ {y}\: '&=3.a.x^{3-1}\\ &=3.a.x^{2} \end{aligned}&\begin{aligned}y&=4x^{4}-x^{2}+2017\\ {y}\: '&=4.4x^{4-1}-2.x^{2-1}+0\\ &=16x^{3}-2x \end{aligned}&\begin{aligned}y&=6-x-3x^{2}\\ {y}\: '&=0-1.x^{1-1}-2.3x^{2-1}\\ &=-1-6x \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \textrm{g}&\textrm{h}&\textrm{i}&\textrm{j}\\\hline \begin{aligned}y&=(x-3)^{2}\\ {y}\: '&=2.(x-3)^{2-1}.1\\ &=2(x-3)\\ &\end{aligned}&\begin{aligned}y&=\left ( x^{3}-2 \right )^{2}\\ {y}\: '&=2.\left ( x^{3}-2 \right )^{2-1}.3x^{2}\\ &=6\left ( x^{3}-2 \right )x^{2}\\ &=6x^{5}-12x^{2} \end{aligned}&\begin{aligned}y&=(x+1)(x-2)\\ {y}\: '&=1.(x+2)+(x+1).1\\ &=2x+3\\ & \end{aligned}&\begin{aligned}y&=(3-x)(5-x)\\ {y}\: '&=-1.(5-x)+(3-x).-1\\ &=2x-8 \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \textrm{k}&\textrm{l}&\multicolumn{2}{|c|}{\textrm{m}}\\\hline \begin{aligned}y&=(2x+3)^{2}\\ {y}\: '&=2.(2x+3)^{2-1}.2\\ &=4(2x+3)^{1}\\ &=8x+12\\ &\\ & \end{aligned}&\begin{aligned}y&=(x-2)^{3}\\ {y}\: '&=3(x-2)^{3-1}.1\\ &=3(x-2)^{2}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=(4x+1)(4x-1)\\ \textrm{c}&\textrm{ara 1}\\ {y}\: '&=4(4x-1)+(4x+1).4\\ &=16x-4+16x+4\\ &=32x\\ & \end{aligned}&\begin{aligned}y&=(4x+1)(4x-1)\\ \textrm{c}&\textrm{ara 2}\\ y&=16x^{2}-1\\ {y}\: '&=2.16x^{2-1}-0\\ &=32x^{1}\\ &=32x \end{aligned}\\\hline \textrm{n}&\textrm{o}&\textrm{p}&\textrm{q}\\\hline \begin{aligned}y&=(x-1)(x+1)(x+2)\\ &=(x^{2}-1)(x+2)\\ &=x^{3}+2x^{2}-x-2\\ {y}\: '&=3x^{3-1}+2.2x^{2-1}-x^{1-1}-0\\ &=3x^{2}+4x-1\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{2}x^{-4}\\ {y}\: '&=-4.\displaystyle \frac{1}{2}x^{-4-1}\\ &=-2x^{-5}\\ &=-\displaystyle \frac{2}{x^{5}}\\ & \end{aligned}&\begin{aligned}y&=x^{-5}\\ {y}'\: &=-5.x^{-5-1}\\ &=-5x^{-6}\\ &=-\displaystyle \frac{5}{x^{6}}\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{2}{x^{3}}\\ &=2x^{-3}\\ {y}\: '&=-3.2x^{-3-1}\\ &=-6x^{-4}\\ &=-\displaystyle \frac{6}{x^{4}} \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \textrm{r}&\textrm{s}&\textrm{t}&\textrm{u}\\\hline \begin{aligned}y&=\displaystyle \frac{1}{2\sqrt{x}}\\ &=\displaystyle \frac{1}{2x^{\frac{1}{2}}}\\ &=\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}\\ {y}\: '&=-\displaystyle \frac{1}{2}.\frac{1}{2}x^{^{-\frac{1}{2}-1}}\\ &=-\displaystyle \frac{1}{4}x^{^{-\frac{3}{2}}}\\ &=-\displaystyle \frac{1}{4x^{^{\frac{3}{2}}}}\\ &=-\displaystyle \frac{1}{4}\sqrt{x^{3}} \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{3x^{5}}\\ &=\displaystyle \frac{1}{3}x^{-5}\\ {y}\: '&=-5.\displaystyle \frac{1}{3}x^{-5-1}\\ &=-\displaystyle \frac{5}{3}x^{-6}\\ &=-\displaystyle \frac{5}{3x^{6}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=2x^{4}+\displaystyle \frac{1}{2x^{3}}\\ &=2x^{4}+\displaystyle \frac{1}{2}x^{-3}\\ {y}\: '&=4.2x^{4-1}+(-3).\displaystyle \frac{1}{2}x^{-3-1}\\ &=8x^{3}-\displaystyle \frac{3}{2}x^{-4}\\ &=8x^{3}-\displaystyle \frac{3}{2x^{4}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{x}{2}+\frac{2}{x}\\ &=\displaystyle \frac{1}{2}x+2x^{-1}\\ {y}\: '&=\displaystyle \frac{1}{2}x^{1-1}+(-1).2x^{-1-1}\\ &=\displaystyle \frac{1}{2}x^{0}-2x^{-2}\\ &=\displaystyle \frac{1}{2}-\displaystyle \frac{2}{x^{2}}\\ &\\ &\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \textrm{v}&\textrm{w}\\\hline \begin{aligned}y&=\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )^{2}\\ &=\left ( 2x^{3}+x^{-1} \right )^{2}\\ {y}\: '&=2.\left ( 2x^{3}+x^{-1} \right )^{2-1}.\left ( 3.2x^{3-1}+(-1)x^{-1-1} \right )\\ &=2.\left ( 2x^{3}+x^{-1} \right )^{1}.\left ( 6x^{2}-x^{-2} \right )\\ &=2\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )\left ( 6x^{2}-\displaystyle \frac{1}{x^{2}} \right )\\ &=2\left ( 12x^{5}-2x+6x-\displaystyle \frac{1}{x^{3}} \right )\\ &=24x^{5}+8x-\displaystyle \frac{2}{x^{3}}\\ & \end{aligned}&\begin{aligned}y&=x^{2}\left ( 1+\sqrt{x} \right )^{2}\\ &=x^{2}\left ( 1+x^{^{\frac{1}{2}}} \right )^{2}\\ {y}\: '&=2x.\left ( 1+x^{^{\frac{1}{2}}} \right )^{2}+x^{2}.2.\left ( 1+x^{^{\frac{1}{2}}} \right )^{2-1}.\left ( 0+\displaystyle \frac{1}{2}.x^{^{\frac{1}{2}-1}} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+2x^{2}.\left ( 1+\sqrt{x} \right ).\left ( \displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+2x^{2}.\left ( \displaystyle \frac{1}{2x^{^{\frac{1}{2}}}} \right ).\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+x^{^{2-\frac{1}{2}}}.\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+x^{^{\frac{3}{2}}}\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+\left ( x\sqrt{x}+x^{2} \right ) \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \multicolumn{2}{|c|}{\textrm{x}}\\\hline \textrm{Cara Pertama}&\textrm{Cara Kedua}\\\hline \begin{aligned}y&=\displaystyle \frac{U}{V}\\ {y}\: '&=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}&\begin{aligned}y&=UV\\ {y}\: '&={U}'.V+U.{V}' \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{aligned}&&&\\ U&=2+4x&\rightarrow {U}'&=4\\ V&=x^{2}&\rightarrow {V}'&=2x\\ &&& \end{aligned}}\\\hline \begin{aligned}y&=\displaystyle \frac{2+4x}{x^{2}}\\ {y}\: '&=\displaystyle \frac{(4)(x^{2})-(2+4x).(2x)}{\left ( x^{2} \right )^{2}}\\ &= \displaystyle \frac{4x^{2}-4x-8x^{2}}{x^{4}}\\ &=\displaystyle \frac{-4x^{2}-4x}{x^{4}}\\ &=\displaystyle \frac{-4x-4}{x^{3}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{2+4x}{x^{2}}\\ &=(2+4x).x^{-2}\\ {y}\: '&=(4).x^{-2}+(2+4x).-2x^{-2-1}\\ &=4x^{-2}-(4+8x).x^{-3}\\ &=4x^{-2}-4x^{-3}-8x^{-2}\\ &=-4x^{-3}-4x^{-2}\\ &=\displaystyle \frac{-4}{x^{3}}+\displaystyle \frac{-4}{x^{2}}\\ &=\displaystyle \frac{-4-4x}{x^{3}}\\ &=\displaystyle \frac{-4x-4}{x^{3}} \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{Ruas kiri = Ruas kanan}}\\\hline \end{array}.

\begin{array}{|l|}\hline \begin{aligned}&\\ &\begin{array}{|r|l|}\hline \textbf{Perhatikanlah bukti poin 1. g}&\\\cline{1-1} \multicolumn{2}{|l|}{.}\\ \multicolumn{2}{|c|}{\LARGE\textrm{\textbf{Cara lain :}}}\\ \multicolumn{2}{|r|}{.}\\\cline{2-2} &\begin{aligned}f(x)&=\tan x=\displaystyle \frac{\sin x}{\cos x}\\\\ &=\displaystyle \frac{U}{V}\quad \Rightarrow \quad \begin{cases} U=\sin x & \rightarrow {U}'=\cos x \\ V=\cos x & \rightarrow {V}'=- \sin x \end{cases}\\\\ {f}\: '(x)&=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}}\\ &=\displaystyle \frac{\cos x.\cos x-\sin x.(-\sin x)}{\left (\cos x \right )^{2}}\\ &=\displaystyle \frac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x\qquad\quad \blacksquare \end{aligned}\\\hline \end{array}\\ & \end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah turunannya}\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=2\sin x\cos x&\textrm{f}.&f(x)=4\sin \left ( 5x+\pi \right )^{3}&\textrm{k}.&f(x)=\sin \left ( \displaystyle \frac{x+3}{x-2} \right )\\ \textrm{b}.&f(x)=x^{2}.\sin x&\textrm{g}.&f(x)=\cos ^{3}(x+5)&\textrm{l}.&f(x)=\cos \left ( x^{2}-6x \right )\\ \textrm{c}.&f(x)=3\cos ^{2}x&\textrm{h}.&f(x)=\sin ^{2}\left ( \pi -3x \right )&\textrm{m}.&f(x)=\displaystyle \frac{1}{\sin x}\\ \textrm{d}.&f(x)=3\sin ^{2}x-x^{3}&\textrm{i}.&f(x)=\displaystyle \frac{\sin ^{2}x}{\cos x}&\textrm{n}.&f(x)=\displaystyle \frac{\sin x}{1+\cos x}\\ \textrm{e}.&f(x)=5\sin ^{4}x&\textrm{j}.&f(x)=\displaystyle \frac{\sin x^{4}}{x^{2}}&\textrm{o}.&f(x)=\displaystyle \frac{\sin x-\cos x}{\sin x+\cos x}\\ \end{array} \end{array}.

Jawab:

\begin{array}{ll|l}\\ 3.\: \textrm{a}&f(x)=2\sin x\cos x&\qquad \textbf{atau}\\ &\begin{aligned}f(x)&=\sin 2x\\ {f}\: '(x)&=\cos 2x.(2)\\ &=2\cos 2x\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\sin 2x\quad \rightarrow \quad \begin{cases} y=\sin u & \rightarrow \displaystyle \frac{dy}{du}=\cos u\\\\ u=2x & \rightarrow \displaystyle \frac{du}{dx}=2 \end{cases} \\ \displaystyle \frac{dy}{dx}&=\displaystyle \frac{dy}{du}.\frac{du}{dx}\\ &=\cos u. 2\\ &=2\cos u\\ &=2\cos 2x \end{aligned} \end{array}.

\begin{array}{ll|l}\\ 3.\: \textrm{h}&f(x)=\sin ^{2}\left ( \pi -3x \right )&\textbf{atau}\\ &\begin{aligned}{f}\: '(x)&=2\sin ^{2-1}\left ( \pi -3x \right ).\cos \left ( \pi -3x \right ).(-3)\\ &=-6\sin \left ( \pi -3x \right )\cos \left ( \pi -3x \right )\\ &=-3.2\sin \left ( \pi -3x \right )\cos \left ( \pi -3x \right )\\ &=-3\sin 2\left ( \pi -3x \right )\\ &=-3\sin \left ( 2\pi -6x \right )\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\sin ^{2}\left ( \pi -3x \right )\quad \rightarrow \quad \begin{cases} y=u^{2} &\rightarrow \displaystyle \frac{dy}{du}=2u. \\\\ u=\sin w &\rightarrow \displaystyle \frac{du}{dw}=\cos w \\\\ w=\pi -3x &\rightarrow \displaystyle \frac{dw}{dx}=-3 \end{cases}\\ \displaystyle \frac{dy}{dx}&=\displaystyle \frac{dy}{du}.\frac{du}{dw}.\frac{dw}{dx}\\ &=\left ( 2u \right ).\left ( \cos w \right ).\left ( -3 \right )\\ &= -3.\left ( 2\sin w \right ).\left ( \cos w \right )\\ &=-3\sin 2w\\ &=-3\sin \left ( 2\pi -6x \right )\end{aligned} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah nilai dari}\: \: \: \displaystyle \frac{dy}{dx}\\ &\textrm{a}.\quad y=\sqrt{x}\\ &\textrm{b}.\quad y=\sqrt{3+\sqrt{x}}\\ &\textrm{c}.\quad y=\sqrt{3+\sqrt{3+\sqrt{x}}}\\ &\textrm{d}.\quad y=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{x}}}} \end{array}\\\\\\ \textrm{Jawab:}\\\\ \begin{array}{|c|c|}\hline \begin{aligned}\textrm{a}.\qquad y&=\sqrt{x}\\ &=x^{^{\frac{1}{2}}}\\ {y}'&=\displaystyle \frac{1}{2}.x^{^{\frac{1}{2}-1}}\\ &=\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}\\ &=\displaystyle \frac{1}{2x^{^{\frac{1}{2}}}}\\ &=\displaystyle \frac{1}{2\sqrt{x}}\\ &\\ & \end{aligned}&\begin{aligned} \textrm{b}.\qquad y&=\sqrt{3+\sqrt{x}}\\ &=\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}}}\\ {y}'&=\displaystyle \frac{1}{2}\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}-1}}.\displaystyle \frac{1}{2}x^{^{\frac{1}{2}-1}}\\ &=\displaystyle \frac{1}{4}\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{-\frac{1}{2}}}.x^{^{-\frac{1}{2}}}\\ &=\displaystyle \frac{1}{4\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}}}.x^{^{\frac{1}{2}}}}\\ &=\displaystyle \frac{1}{4\sqrt{3+\sqrt{x}}.\sqrt{x}}\\ & \end{aligned}\\\hline \end{array}.

C. Turunan Kedua

\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Suatu fungsi}\qquad &\\ &f\: :\: x\: \rightarrow \: y\\ & \end{aligned}}\\\hline \textrm{Notasi}&\textrm{Proses}&\textrm{Contoh Soal}\\\hline \begin{aligned}&\\ &\begin{cases} {f}\: ' (x)& =\displaystyle \frac{dy}{dx} \\\\ {f}\: ''(x) & =\displaystyle \frac{d^{2}y}{dx^{2}} \end{cases}\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ y&=ax^{n}\\ {y}'&=n.a.x^{n-1}\\ {y}''&=(n-1).n.a.x^{n-2}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Misalkan}\: \: y=15x^{3}-4x\\ &\textrm{maka},\\ &\textrm{turunan pertamanya adalah}\quad \displaystyle \frac{dy}{dx}=45x^{2}-4,\: \: \textrm{dan}\\ &\textrm{turunan keduanya adalah}\quad \displaystyle \frac{d^{2}y}{dx^{2}}=90x\\ &\end{aligned}\\\hline \end{array}.

Sebagai tambahan contoh yang lain adalah sebagai berikut:

\begin{array}{|l|}\hline \begin{aligned}&\\ &\begin{aligned}y&=\displaystyle \frac{x^{4}}{3}+\frac{x^{2}}{2}+x+\sqrt{x}+1+\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^{2}}+\frac{3}{x^{4}}\\ &\qquad\quad \parallel \\ &\textbf{Turunan pertama}\\ &\qquad\quad \Downarrow \\ \displaystyle \frac{dy}{dx}={y}\, '&=\displaystyle \frac{4x^{3}}{3}+x+1+\displaystyle \frac{1}{2\sqrt{x}}+0-\frac{1}{2x\sqrt{x}}-\frac{1}{x^{2}}-\frac{2}{x^{3}}-\frac{12}{x^{5}}\\ &=\displaystyle \frac{4x^{3}}{3}+x+1+\displaystyle \frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}-\frac{1}{x^{2}}-\frac{2}{x^{3}}-\frac{12}{x^{5}}\\ &\qquad\quad \parallel \\ &\textbf{Turunan kedua}\\ &\qquad\quad \Downarrow \\ \displaystyle \frac{d^{2}y}{dx^{2}}={y}\, ''&=4x^{2}+1+0-\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^{2}\sqrt{x}}+\frac{2}{x^{3}}+\frac{6}{x^{4}}+\frac{60}{x^{6}}\\ &=4x^{2}+1-\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^{2}\sqrt{x}}+\frac{2}{x^{3}}+\frac{6}{x^{4}}+\frac{60}{x^{6}} \end{aligned}\\ & \end{aligned}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

  1. Kerjakanlah Soal-soal yang belum dijawab/dibahas pada contoh soal
  2. Carilah turunan kedua dari setiap fungsi pada contoh soal No. 2 di atas
  3. Carilah turunan kedua dari setiap fungsi pada contoh soal No. 3 di atas

Sumber Referensi

  1. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (WAJIB). Bandung: SEWU.
  2. Kartini, Suprapto, Subandi, dan Setiyadi, U. (2005). Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
  3. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar Isi 2006. Jakarta: ESIS.
  4. Sunardi, Waluyo, S., Sutrisno, Subagya. 2004. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: Bumi Aksara.
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Lanjutan Contoh Soal Limit Fungsi

\begin{array}{ll}\\ \fbox{11}.&\textrm{Nilai}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( 2x^{2} \right )}{x^{2}+\sin ^{2}3x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{2}{3}&&\textrm{d}.\quad \displaystyle 0\\\\ \textrm{b}.\quad \displaystyle 5\quad &\textrm{c}.\quad \displaystyle \frac{3}{2}\quad &\textrm{e}.\quad \displaystyle \frac{1}{5} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{e}\\\\ \begin{aligned}\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( 2x^{2} \right )}{x^{2}+\sin ^{2}3x}&=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( 2x^{2} \right )}{x^{2}+\sin ^{2}3x}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\sin \left ( 2x^{2} \right )}{x^{2}}}{\displaystyle \frac{x^{2}+\sin ^{2}3x}{x^{2}}}\\ &=\displaystyle \frac{\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( 2x^{2} \right )}{x^{2}}}{\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}}{x^{2}}+\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin ^{2}3x}{x^{2}}}\\ &=\displaystyle \frac{2}{1+3\times 3}\\ &=\displaystyle \frac{2}{10}\\ &=\displaystyle \frac{1}{5} \end{aligned}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Nilai}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 4x\tan ^{2}3x+6x^{3}}{2x^{2}\sin 3x\cos 2x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 0&&\textrm{d}.\quad \displaystyle 5\\\\ \textrm{b}.\quad \displaystyle 3\quad &\textrm{c}.\quad \displaystyle 4\quad &\textrm{e}.\quad \displaystyle 7 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{e}\\\\ \begin{aligned}\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 4x\tan ^{2}3x+6x^{3}}{2x^{2}\sin 3x\cos 2x}&=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 4x\tan ^{2}3x+6x^{3}}{x^{2}.2\sin 3x\cos 2x}\\ &=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 4x\tan ^{2}3x+6x^{3}}{x^{2}\left ( \sin 5x+\sin x \right )}\times \frac{\displaystyle \frac{1}{x^{3}}}{\displaystyle \frac{1}{x^{3}}}\\ &=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\sin 4x\tan ^{2}3x+6x^{3}}{x^{3}}}{\displaystyle \frac{x^{2}\left ( \sin 5x+\sin x \right )}{x^{3}}}\\ &=\displaystyle \frac{\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 4x\tan 3x\tan 3x}{x^{3}}+\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{6x^{3}}{x^{3}}}{\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}\sin 5x}{x^{3}}+\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}\sin x}{x^{3}}}\\ &=\displaystyle \frac{4\times 3\times 3+6}{5+1}\\ &=\displaystyle \frac{42}{6}\\ &=7 \end{aligned}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Nilai}\: \: \underset{x\rightarrow p }{\textrm{Lim}}\: \: \displaystyle \frac{3(x-p)}{\sin (x-p)+2x-2p}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1&&\textrm{d}.\quad \displaystyle 0\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle \frac{1}{3}\quad &\textrm{e}.\quad \displaystyle -1 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}\underset{x\rightarrow p }{\textrm{Lim}}\: \: \displaystyle \frac{3(x-p)}{\sin (x-p)+2x-2p}&=\underset{x\rightarrow p }{\textrm{Lim}}\: \: \displaystyle \frac{3(x-p)}{\sin (x-p)+2x-2p}\times \displaystyle \frac{\displaystyle \frac{1}{(x-p)}}{\displaystyle \frac{1}{(x-p)}}\\ &=\underset{x\rightarrow p }{\textrm{Lim}}\: \: \displaystyle \frac{3}{\displaystyle \frac{\sin (x-p)}{(x-p)}+2}\\ &=\displaystyle \frac{\underset{x\rightarrow p }{\textrm{Lim}}\: \: \displaystyle 3}{\underset{x\rightarrow p }{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x-p)}{(x-p)}+\underset{x\rightarrow p }{\textrm{Lim}}\: \: \displaystyle 2}\\ &=\displaystyle \frac{3}{1+2}\\ &=1 \end{aligned}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Nilai}\: \: \underset{x\rightarrow 1 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{3}-(a+1)x^{2}+ax}{\left ( x^{2}-a \right )\tan (x-1)}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1&&\textrm{d}.\quad \displaystyle 0\\\\ \textrm{b}.\quad \displaystyle 1-a\quad &\textrm{c}.\quad \displaystyle a\quad &\textrm{e}.\quad \displaystyle 2-a \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}\underset{x\rightarrow 1 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{3}-(a+1)x^{2}+ax}{\left ( x^{2}-a \right )\tan (x-1)}&=\underset{x\rightarrow 1 }{\textrm{Lim}}\: \: \displaystyle \frac{x(x-a)(a-1)}{\left ( x^{2}-a \right )\tan (x-1)}\\ &=\underset{x\rightarrow 1 }{\textrm{Lim}}\: \: \displaystyle \frac{x(x-a)(x-1)}{\left ( x^{2}-a \right )\tan (x-1)}\times \displaystyle \frac{\displaystyle \frac{1}{(x-1)}}{\displaystyle \frac{1}{(x-1)}}\\ &=\underset{x\rightarrow 1 }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x(x-a)(x-1)}{(x-1)}}{\displaystyle \frac{\left ( x^{2}-a \right )\tan (x-1)}{(x-1)}}\\ &=\underset{x\rightarrow 1 }{\textrm{Lim}}\: \: \displaystyle \frac{x(x-a)}{\left ( x^{2}-a \right )}\\ &=\displaystyle \frac{1(1-a)}{(1-a)}\\ &=1 \end{aligned}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Nilai}\: \: \underset{x\rightarrow -3 }{\textrm{Lim}}\: \: \displaystyle \frac{1-\cos (x+3)}{x^{2}+6x+9}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2&&\textrm{d}.\quad -\displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad \displaystyle -2\quad &\textrm{c}.\quad \displaystyle \frac{1}{2}\quad &\textrm{e}.\quad \displaystyle \frac{1}{3} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}\underset{x\rightarrow -3 }{\textrm{Lim}}\: \: \displaystyle \frac{1-\cos (x+3)}{x^{2}+6x+9}&=\underset{x\rightarrow -3 }{\textrm{Lim}}\: \: \displaystyle \frac{1-\cos (x+3)}{(x+3)^{2}}\\ &=\underset{x\rightarrow -3 }{\textrm{Lim}}\: \: \displaystyle \frac{2\sin ^{2}\displaystyle \frac{(x+3)}{2}}{(x+3)^{2}}\\ &=\underset{x\rightarrow -3 }{\textrm{Lim}}\: \: \displaystyle \frac{2\sin \displaystyle \frac{1}{2}(x+3)}{(x+3)}\times \underset{x\rightarrow -3 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin \displaystyle \frac{1}{2}(x+3)}{(x+3)}\\ &=2\times \displaystyle \frac{1}{2}\times \frac{1}{2}\\ &=\displaystyle \frac{1}{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{16}.&\textbf{(SNMPTN Mat IPA 2012)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{1-\cos ^{2}x}{x^{2}\tan \left ( x+\displaystyle \frac{\pi }{3} \right )}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \sqrt{3}&&\textrm{d}.\quad \displaystyle \frac{\sqrt{3}}{2}\\\\ \textrm{b}.\quad \displaystyle 0\quad &\textrm{c}.\quad \displaystyle \frac{\sqrt{3}}{3}\quad &\textrm{e}.\quad \displaystyle \sqrt{3}\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}\underset{x\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{1-\cos ^{2}x}{x^{2}\tan \left ( x+\displaystyle \frac{\pi }{3} \right )}&=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin ^{2}x}{x^{2}\tan \left ( x+\displaystyle \frac{\pi }{3} \right )}\\ &=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sin ^{2}x}{x^{2}}\times \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{\tan \left ( x+\displaystyle \frac{\pi }{3} \right )}\\ &=1\times \frac{1}{\tan \left ( 0+\displaystyle \frac{\pi }{3} \right )}\\ &=\displaystyle \frac{1}{\tan \displaystyle \frac{\pi }{3}}\\ &=\cot \displaystyle \frac{\pi }{3}\\ &=\displaystyle \frac{\sqrt{3}}{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-2,\: \: \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle x^{2}-2&&\textrm{d}.\quad \displaystyle x\\\\ \textrm{b}.\quad \displaystyle x^{2}\quad &\textrm{c}.\quad \displaystyle 2x\quad &\textrm{e}.\quad \displaystyle 2x-2 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}\textrm{Diketahui bahwa}\: \: \: f(x)&=x^{2}-2,\\ \textrm{maka nilai untuk}\qquad \: \, \, &\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\, \, \, \, &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\left ((x+h)^{2}-2 \right )-\left ( x^{2}-2 \right )}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{x^{2}+2xh+h^{2}-2-x^{2}+2}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{2xh+h^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \left (2x+h \right )\\ &=2x \end{aligned}.

\begin{array}{ll}\\ \fbox{18}.&\textrm{Diketahui}\: \: f(x)=\sqrt{x-1}\: ,\: \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{f(2+h)-f(2)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle 1\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle 0\quad &\textrm{e}.\quad \displaystyle -1\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}\textrm{Diketahui bahwa}\: \: \: f(x)&=\sqrt{x-1},\\ \textrm{maka nilai untuk}\qquad \: \, \, &\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(2+h)-f(2)}{h}\: \, \, \, &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\times \displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{(h+1)-1}{h\times \left ( \sqrt{h+1}+1 \right )}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt{h+1}+1 \right )}\\ &=\displaystyle \frac{1}{\sqrt{0+1}+1}\\ &=\displaystyle \frac{1}{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{19}.&\textbf{(Mat Das SIMAK UI 2013)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow 5}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}+\sqrt{2}&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 5-2\sqrt{6}\quad &\textrm{c}.\quad 2\sqrt{6}\quad &\textrm{e}.\quad 5+2\sqrt{6}\end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{e}\\\\ \begin{aligned}\underset{x\rightarrow 5}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}&=\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}\\ &=\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{3+2+2\sqrt{6}}{3-2}\\ &=5+2\sqrt{6} \end{aligned}.

\begin{array}{ll}\\ \fbox{20}.&\textbf{(Mat IPA SBMPTN 2014)}\\\\ &\textrm{Jika}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4\: \: \textrm{dan}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3,\\\\ &\textrm{maka nilai}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x).g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{14}&&\textrm{d}.\quad \displaystyle \frac{4}{14}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{14}\quad &\textrm{c}.\quad \displaystyle \frac{3}{14}\quad &\textrm{e}.\quad \displaystyle \frac{5}{14}\end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{b}\\\\ \begin{aligned}&\textrm{Perhatikan bahwa}\: ,\\ &\begin{array}{lll}\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4&&\\ &&\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3&+&\\ &&\\\cline{1-1} &&\\ 2\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x)\qquad\qquad\: \: =1&&\\ &&\\ \qquad\qquad\: \: \: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x)=\displaystyle \frac{1}{2}\: ,&\textrm{sehingga}&\underset{x\rightarrow a}{\textrm{Lim}}\: \: f(g)=\displaystyle \frac{2}{7}\\ &&\\ \textrm{maka}\: ,&&\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x).g(x)=\displaystyle \frac{1}{2}\times \frac{2}{7}=\displaystyle \frac{2}{14}&&\\\end{array} \end{aligned}.

Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Setiyadi, U. (2005). Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
  2. Suparmin, S. dan Malau, A. (2014). Mainstream Matematika Dasar & Matematika IPA untuk Siswa SMA/MA Kelompok MIA. Bandung: Yrama Widya.
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Contoh Soal Limit Fungsi

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle \frac{1}{4}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{4}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{2}\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&= \left ( \displaystyle \frac{6-2}{2^{2}-4}-\frac{1}{2-2} \right )\\ &=\left ( \displaystyle \frac{4}{0}-\frac{1}{0} \right )=\infty -\infty \\ &\: \: \textrm{hal ini tidak diperkenankan}\\ \textrm{Sehingga},\, \qquad\qquad\qquad &\\ \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{(x+2)}{(x-2)(x+2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{x+2}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{4-2x}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2(x-2)}{(x+2)(x-2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2}{x+2} \right )\\ &=\displaystyle -\frac{2}{(2+2)}\\ &=-\displaystyle \frac{1}{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Nilai}\: \: \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 2&&\textrm{d}.\quad \displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle 2\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&= \left ( \displaystyle \frac{4-4}{2^{4}-4} \right )\\ &=\displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehingga},\: \: \: \: \quad&\\ \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{\sqrt{x}\left ( 2-\sqrt{x} \right )} \\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{-\sqrt{x}\left ( \sqrt{x}-2 \right )}\\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: -\displaystyle \frac{\left ( \sqrt{x}+2 \right )}{\sqrt{x}}\\ &=-\displaystyle \frac{\sqrt{4}+2}{\sqrt{4}}\\ &=-\displaystyle \frac{2+2}{2}\\ &=-2 \end{aligned}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Nilai}\: \: \underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}&&\textrm{d}.\quad \displaystyle 2\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\quad &\textrm{c}.\quad 1\quad &\textrm{e}.\quad \displaystyle 4\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}&= \displaystyle \frac{\left ( 2-3+1 \right )\left ( 1-1 \right )}{\left ( 1-1 \right )^{2}} \\ &=\displaystyle \frac{0\times 0}{0^{2}}\\ &=\displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehingga},\, \, \: \: \: \quad \qquad \qquad \qquad\quad&\\ \underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}&=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \left ( 2\sqrt{x}-1 \right )\times \left ( \sqrt{x}-1 \right ) \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )\left ( \sqrt{x}-1 \right )}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \left ( 2\sqrt{x}-1 \right )\\ &=2.1-1\\ &=2-1\\ &=1 \end{aligned}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Nilai}\: \: \underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{1}{14}\sqrt{7}&&\textrm{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{7}\sqrt{7}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{14}\sqrt{7}\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}&=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\times \displaystyle \frac{\sqrt{x+4}+\sqrt{2x+1}}{\sqrt{x+4}+\sqrt{2x+1}}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x+4 \right )-\left ( 2x+1 \right )}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: -\displaystyle \frac{-x+3}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{-1}{\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=-\displaystyle \frac{1}{\left ( \sqrt{7}+\sqrt{7} \right )}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\times \displaystyle \frac{\sqrt{7}}{\sqrt{7}}\\ &=-\displaystyle \frac{1}{14}\sqrt{7} \end{aligned}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Jika}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6,\: \: \textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2&&\textrm{d}.\quad \displaystyle -2\\\\ \textrm{b}.\quad \displaystyle 1\quad &\textrm{c}.\quad -1\quad &\textrm{e}.\quad \displaystyle -3\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{e}\\\\ \begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}&=6\\ &\: \: \: \textrm{dengan bantuan limit kanan yaitu}\: \: x=2+h\: \Rightarrow \: h\rightarrow 0\\ \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left ( 2+h \right )-2a}{\sqrt{2\left ( 2+h \right )}-\left ( 2+h \right )}&=6\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2a+ah-2a}{\sqrt{4+2h}-\left ( 2+h \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah}{\sqrt{4+2h}-\left ( 2+h \right )}\times \displaystyle \frac{\left ( \sqrt{4+2h}+\left ( 2+h \right ) \right )}{\left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{4+2h-\left ( 4+4h+h^{2} \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2h-h^{2}}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2-h}\\ 6&=\displaystyle \frac{a\times \left (\sqrt{4+0} +\left ( 2+0 \right ) \right )}{-2-0}\\ 6&=\displaystyle \frac{a\left ( \sqrt{4}+2 \right )}{-2}\\ \displaystyle \frac{a(4)}{-2}&=6\\ a(-2)&=6\\ a&=-3 \end{aligned}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 4x-3 \right )^{2}}{\left ( 2x+5 \right )^{2}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 8&&\textrm{d}.\quad \displaystyle 1\\\\ \textrm{b}.\quad \displaystyle 4\quad &\textrm{c}.\quad 2\quad &\textrm{e}.\quad \displaystyle -2\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{b}\\\\ \begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 4x-3 \right )^{2}}{\left ( 2x+5 \right )^{2}}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{16x^{2}-24x+9}{4x^{2}+20x+25}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{16x^{2}}{x^{2}}-\displaystyle \frac{24x}{x^{2}}+\displaystyle \frac{9}{x^{2}}}{\displaystyle \frac{4x^{2}}{x^{2}}+\displaystyle \frac{20x}{x^{2}}+\displaystyle \frac{25}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{16-\displaystyle \frac{24}{x}+\displaystyle \frac{9}{x^{2}}}{4+\displaystyle \frac{20}{x}+\displaystyle \frac{25}{x^{2}}}\\ &=\displaystyle \frac{16-0+0}{4+0+0}\\ &=\displaystyle \frac{16}{4}\\ &=4 \end{aligned}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{2x^{2}+3x}{\sqrt{x^{2}-x}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 0&&\textrm{d}.\quad \displaystyle 2\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle 1\quad &\textrm{e}.\quad \displaystyle \infty \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{e}\\\\ \begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{2x^{2}+3x}{\sqrt{x^{2}-x}}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{2x^{2}}{x^{2}}+\displaystyle \frac{3x}{x^{2}}}{\displaystyle \frac{1}{x^{2}}\sqrt{x^{2}-x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{2x^{2}}{x^{2}}+\displaystyle \frac{3x}{x^{2}}}{\sqrt{\displaystyle \frac{x^{2}}{x^{4}}-\displaystyle \frac{x}{x^{4}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{2+\displaystyle \frac{3}{x}}{\sqrt{\displaystyle \frac{1}{x^{2}}-\displaystyle \frac{1}{x^{3}}}}\\ &=\displaystyle \frac{2+0}{\sqrt{0-0}}\\ &=\displaystyle \frac{2}{0}\\ &=\displaystyle \infty \end{aligned}.

\begin{array}{ll}\\ \fbox{8}.&\textbf{(USM UGM Mat IPA)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{3}&&\textrm{d}.\quad -\displaystyle \frac{2}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\quad &\textrm{c}.\quad -\displaystyle \frac{1}{3}\quad &\textrm{e}.\quad -\displaystyle \frac{5}{3}\end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{e}\\\\.

\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-\sqrt[3]{\left ( x+1 \right )^{3}} \right )\\ &\: \: \: \textrm{ingat bentuk}\: \: a-b=\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )\\ &\: \: \: \textrm{dan untuk}\: \: \begin{cases} a & =\left ( x^{3}-2x^{2} \right ) \\ & \\ b & = \left ( x+1 \right )^{3} \end{cases}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{a-b}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x+1 \right )^{3}}{\sqrt[3]{\left ( x^{3}-2x^{2} \right )^{2}}+\sqrt[3]{\left ( x^{3}-2x^{2} \right )\left ( x+1 \right )^{3}}+\sqrt[3]{\left ( \left ( x+1 \right )^{3} \right )^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x^{3}+3x^{2}+3x+1 \right )}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{-5x^{2}+...}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\displaystyle \frac{-5}{1+1+1}\\ &=-\displaystyle \frac{5}{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Nilai}\: \: \underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+\cdots +\frac{1}{k\times (k+1)} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1&&\textrm{d}.\quad \displaystyle \frac{5}{2}\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\quad &\textrm{c}.\quad \displaystyle 2\quad &\textrm{e}.\quad \displaystyle \infty \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )&=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \left (1-\frac{1}{2} \right )+\left (\frac{1}{2}-\frac{1}{3} \right )+\left (\frac{1}{3}-\frac{1}{4} \right )+\cdots +\left (\frac{1}{k}-\frac{1}{k+1} \right )\right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( 1-\frac{1}{k+1} \right )\\ &=\displaystyle \left ( 1-\frac{1}{\infty +1} \right )\\ &=\displaystyle 1-\frac{1}{\infty }\\ &=1-0\\ &=1 \end{aligned}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Nilai}\: \: \underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\tan x}{2x^{2}+3x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 3&&\textrm{d}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{b}.\quad \displaystyle 1\quad &\textrm{c}.\quad \displaystyle 0\quad &\textrm{e}.\quad -\displaystyle \frac{1}{3} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{d}\\\\ \begin{aligned}\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\tan x}{2x^{2}+3x}&=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\tan x}{x}\times \frac{1}{2x+3}\\ &=\underset{x\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\tan x}{x}\times \underset{x\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{1}{2x+3}\\ &=1\times \underset{x\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{1}{2x+3}\\ &=\displaystyle \frac{1}{0+3}\\ &=\displaystyle \frac{1}{3} \end{aligned}.

 

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Limit Fungsi Trigonometri

Perhatikanlah ilustrsi gambar berikut ini

serta ,

\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\textrm{Luas Daerah}}\\\hline \triangle \textrm{AOB}&\textrm{Juring AOB}&\triangle \textrm{AOC}\\\hline \begin{aligned}\textrm{L}_{_{\triangle \textrm{AOB}}}&=\displaystyle \frac{1}{2}.OA.BD\\ &=\displaystyle \frac{1}{2}.OA.OB.\sin x\\ &=\displaystyle \frac{1}{2}.1.1.\sin x\\ &=\displaystyle \frac{1}{2}\sin x \end{aligned}&\begin{aligned}\textrm{L}_{_{\textrm{Juring AOB}}}&=\displaystyle \frac{x}{2\pi }.\pi r^{2}\\ &=\displaystyle \frac{x}{2\pi }.\pi .1^{2}\\ &=\displaystyle \frac{1}{2}x\\ &\\ & \end{aligned}&\begin{aligned}\textrm{L}_{_{\triangle \textrm{AOB}}}&=\displaystyle \frac{1}{2}.OA.AC\\ &=\displaystyle \frac{1}{2}.OA.OA.\tan x\\ &=\displaystyle \frac{1}{2}.1.1.\tan x\\ &=\displaystyle \frac{1}{2}\tan x \end{aligned}\\\hline \end{array}.

Untuk prinsip apit

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{L}_{_{\triangle \textrm{AOB}}}\leq \textrm{L}_{_{\textrm{Juring AOB}}}\leq \textrm{L}_{_{\triangle \textrm{AOB}}}}\\\hline \begin{array}{rlclll}\\ \displaystyle \frac{1}{2}\sin x&\leq &\displaystyle \frac{1}{2}x&\leq &\displaystyle \frac{1}{2}\tan x\\\\ \sin x&\leq &x&\leq &\displaystyle \frac{\sin x}{\cos x}\\\\ 1&\leq &\displaystyle \frac{x}{\sin x}&\leq &\displaystyle \frac{1}{\cos x}\\\\ 1&\geq &\displaystyle \frac{\sin x}{x}&\geq &\cos x\\\\ \cos x&\leq &\displaystyle \frac{\sin x}{x}&\leq &1\\\\ &&&& \end{array}&\begin{array}{rlclll}\\ \displaystyle \frac{1}{2}\sin x&\leq &\displaystyle \frac{1}{2}x&\leq &\displaystyle \frac{1}{2}\tan x\\\\ \displaystyle \sin x&\leq &x&\leq &\displaystyle \tan x\\\\ \displaystyle \frac{\sin x}{\tan x}&\leq &\displaystyle \frac{x}{\tan x}&\leq &1\\\\ \cos x&\leq &\displaystyle \frac{ x}{\tan x}&\leq &1\\\\ \displaystyle \frac{1}{\cos x}&\geq &\displaystyle \frac{\tan x}{x}&\geq &1\\ 1&\leq &\displaystyle \frac{\tan x}{x}&\leq &\displaystyle \frac{1}{\cos x} \end{array} \\\hline \end{array}.

\begin{aligned}&\textrm{Jika}\: \: x\rightarrow 0,\: \textrm{maka}\: \: \underset{x\rightarrow 0}{\textrm{lim}}\: \cos x=1,\: \textrm{akan diperoleh}\: \:\begin{cases} \bullet & \underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}=1\\\\ \bullet & \underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}=1 \\\\ \bullet & \underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x}{\tan x}=1 \\\\ \bullet & \underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\tan x}{x}=1 \end{cases} \end{aligned}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{aligned}1.\quad \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 4x}{\sin 7x}&=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 4x}{\sin 7x} \\ &=\underset{x\rightarrow 0 }{\textrm{lim}}\: \left ( \displaystyle \frac{\tan 4x}{4x} \right )\left ( \displaystyle \frac{7x}{\sin 7x} \right )\left ( \displaystyle \frac{4x}{7x} \right )\\ &=1.1.\displaystyle \frac{4}{7}\\ &=\displaystyle \frac{4}{7}\\ \end{aligned}.

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