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.

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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