Lanjutan Limit Fungsi

B. Sifat-Sifat Limit Fungsi Aljabar

\begin{array}{ll}\\ 1.&\underset{x\rightarrow a}{\textrm{lim}}\: k=k\\ 2.&\underset{x\rightarrow a}{\textrm{lim}}\: x=a\\ 3.&\underset{x\rightarrow a}{\textrm{lim}}\: k.f(x)=k.\underset{x\rightarrow a}{\textrm{lim}}\: f(x)\\ 4.&\underset{x\rightarrow a}{\textrm{lim}}\: \left ( f(x)+g(x) \right )=\underset{x\rightarrow a}{\textrm{lim}}\: f(x)+\underset{x\rightarrow a}{\textrm{lim}}\: g(x)\\ 5.&\underset{x\rightarrow a}{\textrm{lim}}\: \left ( f(x)-g(x) \right )=\underset{x\rightarrow a}{\textrm{lim}}\: f(x)-\underset{x\rightarrow a}{\textrm{lim}}\: g(x)\\ 6.&\underset{x\rightarrow a}{\textrm{lim}}\: \left ( f(x)\times g(x) \right )=\underset{x\rightarrow a}{\textrm{lim}}\: f(x)\times \underset{x\rightarrow a}{\textrm{lim}}\: g(x)\\ 7.&\underset{x\rightarrow a}{\textrm{lim}}\: \displaystyle \frac{f(x)}{g(x)}=\displaystyle \frac{\underset{x\rightarrow a}{\textrm{lim}}\: f(x)}{\underset{x\rightarrow a}{\textrm{lim}}\: g(x)},\quad \textrm{dengan}\: \: \: g(x)\neq 0\\ 8.&\underset{x\rightarrow a}{\textrm{lim}}\: \left (f(x) \right )^{n}= \left [\underset{x\rightarrow a}{\textrm{lim}}\: f(x)) \right ]^{n}\\ 9.&\underset{x\rightarrow a}{\textrm{lim}}\: \sqrt[n]{f(x)}=\sqrt[n]{\underset{x\rightarrow a}{\textrm{lim}}\: f(x)},\quad \textrm{dengan}\: \: \underset{x\rightarrow a}{\textrm{lim}}\: f(x)\geq 0,\: \textrm{saat}\: \: n\: \: \textrm{genap} \end{array}.

Langkah penyelesaian limit fungsi aljabar adalah sebagaimana berikut ini:

  • Substitusi langsung
  • Faktorisasi
  • Mengalikan dengan faktor sekawan

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

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Substitusi Langsung}}\\\hline \begin{aligned}&\\ \underset{x\rightarrow -1}{\textrm{lim}}\: x^{2}+3&=(-1)^{2}+3\\ &=1+3\\ &=4\\ & \end{aligned}&\begin{aligned}\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x^{3}+1}{x+1}&=\displaystyle \frac{0^{3}+1}{0+1}\\ &=\displaystyle \frac{1}{1}\\ &=1 \end{aligned}\\\hline \end{array}.

Perhatikan juga contoh soal berikut!

\begin{aligned}&\textrm{Diberikan polinom}\: \: f(x)=a_{_{n}}x^{^{n}}+a_{_{n-1}}x^{^{n-1}}+a_{_{n-2}}x^{^{n-2}}+...+a_{_{2}}x^{^{2}}+a_{_{1}}x+a_{_{0}}.\\ &\textrm{Tunjukkan bahwa}\: \: \underset{x\rightarrow a}{\textrm{lim}}\: f(x)=f(a) \end{aligned}\\\\\\ \begin{aligned}\textbf{Bukti}:\, \, \, &\\ &\\ \underset{x\rightarrow a}{\textrm{lim}}\: f(x)&=\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{n}}x^{^{n}}+a_{_{n-1}}x^{^{n-1}}+a_{_{n-2}}x^{^{n-2}}+...+a_{_{2}}x^{^{2}}+a_{_{1}}x^{1}+a_{_{0}}\\ &=\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{n}}x^{^{n}}+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{n-1}}x^{^{n-1}}+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{n-2}}x^{^{n-2}}+...+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{2}}x^{^{2}}+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{1}}x^{^{1}}+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{0}}\\ &=a_{n}\underset{x\rightarrow a}{\textrm{lim}}\: x^{n}+a_{n-1}\underset{x\rightarrow a}{\textrm{lim}}\: x^{n-1}+a_{n-2}\underset{x\rightarrow a}{\textrm{lim}}\: x^{n-2}+...+ a_{2}\underset{x\rightarrow a}{\textrm{lim}}\: x^{2}+a_{1}\underset{x\rightarrow a}{\textrm{lim}}\: x^{1}+a_{0}\\ &=a_{n}\left ( \underset{x\rightarrow a}{\textrm{lim}}\: x \right )^{n}+a_{n-1}\left ( \underset{x\rightarrow a}{\textrm{lim}}\: x \right )^{n-1}+a_{n-2}\left ( \underset{x\rightarrow a}{\textrm{lim}}\: x \right )^{n-2}+...+a_{2}\left ( \underset{x\rightarrow a}{\textrm{lim}}\: x \right )^{2}+a_{1}\underset{x\rightarrow a}{\textrm{lim}}\: x+a_{0}\\ &=a_{n}a^{n}+a_{n-1}a^{n-1}+a_{n-2}a^{n-2}+...+a_{2}a^{2}+a_{1}a+a_{0}\\ &=f(a)\qquad \blacksquare \end{aligned}.

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Pemfaktoran}}\\\hline \begin{aligned}\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{2}-4}{x-2}&=\displaystyle \frac{4-4}{2-2}=\frac{0}{0}\\ &\textrm{hasil seperti ini}\\ &\textrm{harus dihindari, maka}\\ \underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{2}-4}{x-2}&=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{(x-2)(x+2)}{(x-2)}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: (x+2)\\ &=2+2\\ &=4\\ & \end{aligned}&\begin{aligned}\underset{t\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{t^{3}-8}{t^{2}+t-6}&=\displaystyle \frac{8-8}{4+2-6}=\frac{0}{0}\\ &\textrm{hasil seperti ini lagi-lagi}\\ &\textrm{harus dihindari, maka}\\ \underset{t\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{t^{3}-8}{t^{2}+t-6}&=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{(t-2)(t^{2}+2t+4)}{(t-2)(t+3)}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{t^{2}+2t+4}{t+3}\\ &=\displaystyle \frac{2^{2}+2.2+4}{2+3}\\ &=\displaystyle \frac{12}{5} \end{aligned} \\\hline \end{array}.

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Mengalikan dengan faktor sekawan}}\\\hline \begin{aligned}\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x-4}{\sqrt{x}-2}&=\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x-4}{\sqrt{x}-2}\times \displaystyle \frac{\sqrt{x}+2}{\sqrt{x}+2}\\ &=\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{(x-4)\left ( \sqrt{x}+2 \right )}{x-4}\\ &=\underset{x\rightarrow 0}{\textrm{lim}}\: \left (\sqrt{x}+2 \right )\\ &=\sqrt{0}+2\\ &=2\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{\sqrt{x+2}-\sqrt{6-x}}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{\sqrt{x+2}-\sqrt{6-x}}{x-2}\times \left ( \displaystyle \frac{\sqrt{x+2}+\sqrt{6-x}}{\sqrt{x+2}+\sqrt{6-x}} \right )\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{(x+2)-(6-x)}{(x-2)\left ( \sqrt{x+2}+.... \right )}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{2(x-2)}{(x-2)\left ( \sqrt{x+2}+.... \right )}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{....}{\sqrt{x+2}+....}\\ &=\displaystyle \frac{....}{\sqrt{4}+\sqrt{....}}\\ &=\displaystyle \frac{....}{2+....}\\ &=\displaystyle \frac{....}{....}\\ & \end{aligned} \\\hline \end{array}.

Berikut soal aplikasi untuk model pemfaktoran, yaitu:

\begin{aligned}&\textrm{Tentukanlah nilai}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{jika}\: \: \underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{ax+b}{x^{4}-81}=5\\ &\textrm{Jawab}: \end{aligned}\\\\\\ \begin{array}{|l|l|}\hline \begin{aligned}\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{ax+b}{x^{4}-81}&=5\\ \bullet \quad \displaystyle \frac{f(3)}{g(3)}&=\frac{0}{0}\Leftrightarrow \displaystyle \frac{3a+b}{3^{4}-81}=\frac{0}{0}\\ \Rightarrow b&=-3a\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{Dengan bantuan}&\: \textrm{limit \textsl{kanan} kita mendapatkan}\\ \textrm{dengan}\: \: \: \: x=3+&h\: \: \Rightarrow \: \: h\rightarrow 0,\: \textrm{maka}\\ \bullet \quad \underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{ax+b}{x^{4}-81}&=\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{a(3+h)+b}{(3+h)^{4}-81}=5\\ 5&=\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{3a+ah+b}{h^{4}+12h^{3}+54h^{2}+108h+81-81}\\ 5&=\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{3a+ah+(-3a)}{h^{4}+12h^{3}+54h^{2}+108h}\\ 5&=\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{a}{h^{3}+12h^{2}+54h+108}\\ 5&=\displaystyle \frac{a}{0+0+0+108}\\ 108\times 5&=a\\ a&=540\\ \textrm{sehingga},\quad &\\ b&=-3a\\ &=-3(540)\\ &=-1620 \end{aligned}\\\hline \end{array}.

Sebagai pembanding untuk mengecek jawaban di atas adalah sebagai berikut:

\begin{aligned}\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{ax+b}{x^{4}-81}&=\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{540x-1620}{x^{4}-81}\: \: ,\qquad\quad \textrm{ingat}\: \: \begin{cases} a & =540 \\ \textrm{dan}&\\ b & =-1620 \end{cases}\\ &=\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{540(x-3)}{(x^{2}+9)(x^{2}-9)}\\ &=\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{540(x-3)}{(x^{2}+9)(x+3)(x-3)}\\ &=\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{540}{(x^{2}+9)(x+3)}\\ &=\displaystyle \frac{540}{(3^{2}+9)(3+3)}\\ &=\displaystyle \frac{540}{(18)(6)}\\ &=5 \end{aligned}.

soal aplikasi lainnya untuk model pemfaktoran:

\begin{aligned}&\textrm{Tentukanlah harga}\: \: n\in \mathbb{N}\: ,\: \textrm{jika}\: \: \: \underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{n}-2^{n}}{x-2}=192\end{aligned}\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{n}-2^{n}}{x-2}&=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{(x-2)\left ( x^{n-1}+x^{n-2}.2+x^{n-3}.2^{2}+...+x^{3}.2^{n-4}+x^{2}.2^{n-3}+x.2^{n-2}+2^{n-1} \right )}{(x-2)}\\\\ 192&=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \left (x^{n-1}+x^{n-2}.2+x^{n-3}.2^{2}+...+x^{3}.2^{n-4}+x^{2}.2^{n-3}+x.2^{n-2}+2^{n-1} \right )\\ 192&=2^{n-1}+2^{n-2}.2+2^{n-3}.2^{2}+...+2^{3}.2^{n-4}+2^{2}.2^{n-3}+2.2^{n-2}+2^{n-1}\\ 192&=\underset{\textrm{sebanyak n faktor}}{\underbrace{2^{n-1}+2^{n-1}+2^{n-1}+...+2^{n-1}+2^{n-1}+2^{n-1}+2^{n-1}}}\\ 192&=n.2^{n-1}\\ 6.32&=n.2^{n-1}\\ 6.2^{6-1}&=n.2^{n-1} \end{aligned}\\\\ \textrm{Jadi},\: n=6.

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

Tentukanlah limit-limit berikut!

\begin{array}{lll}\\ 1.&\textrm{a}.&\underset{x\rightarrow 1}{\textrm{lim}}\: (x+1)(x-8)\\ &\textrm{b}.&\underset{x\rightarrow 2}{\textrm{lim}}\: (3x+5)^{2}\\ &\textrm{c}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{\sqrt{x^{2}+5}}{x+2}\\ &\textrm{d}.&\underset{x\rightarrow 1}{\textrm{lim}}\: \left ( \displaystyle \frac{\sqrt{3x^{2}+2x}}{\sqrt{5x-6}}-\frac{x}{2} \right ) \end{array}.

\begin{array}{lll}\\ 2.&\textrm{a}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{3}-8}{x-2}\\ &\textrm{b}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{2x+\sqrt{x}}{\sqrt{x}}\\ &\textrm{c}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{3}+3x^{2}+4}{x^{3}-2x^{2}-4x+8}\\ &\textrm{d}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{2x^{9}-9x^{2}}{5x^{7}-x^{2}} \end{array}.

\begin{array}{lll}\\ 3.&\textrm{a}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sqrt{9+x}-\sqrt{9-x}}{x}\\ &\textrm{b}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sqrt{x^{2}+x}+\sqrt{x}}{x\sqrt{x}}\\ &\textrm{c}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{\sqrt{2x^{2}+5}-\sqrt{x}+7}{\sqrt{3x-4}-\sqrt{x}}\\ &\textrm{d}.&\underset{x\rightarrow 1}{\textrm{lim}}\: \displaystyle \frac{\sqrt[3]{2x+6}-2}{x-1}\\ &\textrm{e}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sqrt{a-x}-\sqrt{a+x}}{\sqrt{b-x}-\sqrt{b+x}} \end{array}.

\begin{array}{lll}\\ 4.&\textrm{a}.&\underset{x\rightarrow -3}{\textrm{lim}}\: \left (\displaystyle \frac{x+6}{x^{2}+5x+6}-\displaystyle \frac{x}{x+3} \right )\\ &\textrm{b}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \left (\displaystyle \frac{x-12}{x^{2}+2x-8}-\displaystyle \frac{x^{2}+1}{3x^{2}-15x+18} \right )\\ \end{array}.

Tentukanlah nilai  a  dan  b berikut!

\begin{array}{lll}\\ 5.&\textrm{a}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sqrt{ax+b}}{x}=\displaystyle 1\\ &\textrm{b}.&\underset{x\rightarrow 4}{\textrm{lim}}\: \displaystyle \frac{ax+b-\sqrt{x}}{x-4}=\displaystyle \frac{3}{4}\\ &\textrm{b}.&\underset{x\rightarrow 4}{\textrm{lim}}\: \displaystyle \frac{ax+b-\sqrt{x}}{x-4}=\displaystyle \frac{1}{2}\\ &\textrm{c}.&\underset{x\rightarrow 4}{\textrm{lim}}\: \displaystyle \frac{ax^{2}+bx-\sqrt{x}}{x^{2}-16}=\displaystyle \frac{1}{2} \end{array}.

Sumber Referensi

  1. Djumanta, W. dan Sudrajat, R. (2008). Mahir Mengembangkan Kemampuan Matematika untuk Sekolah Menengah Atas/Madrasah Aliyah Kelas XI Program Ilmu Pengetahuan Alam. Jakarta: Pusat Perbukuan, Departemen Pendidikan Nasional.
  2. Kartini, Suprapto, Subandi, dan Setiyadi, U. (2005). Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.

 

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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