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}.

 

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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