Contoh Soal 2 Limit Fungsi Aljabar

$\begin{array}{ll}\\ 6.&\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-2,\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle x^{2}-2&&\textrm{d}.\quad \displaystyle x\\\\ \textrm{b}.\quad \displaystyle x^{2}\quad &\textrm{c}.\quad \displaystyle 2x\quad &\textrm{e}.\quad \displaystyle 2x-2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\: f(x)=x^{2}-2,\\ \textrm{maka nila}&\textrm{i untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(x+h)-f(x)}{h}\, \, \, \, \\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\left ((x+h)^{2}-2 \right )-\left ( x^{2}-2 \right )}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{x^{2}+2xh+h^{2}-2-x^{2}+2}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{2xh+h^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \left (2x+h \right )\\ &=\color{red}2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: f(x)=\sqrt{x-1},\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{f(2+h)-f(2)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle 1\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle 0\quad &\textrm{e}.\quad \displaystyle -1\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}\: f(x)=\sqrt{x-1},\\ \textrm{maka}\: \: & \textrm{nilai untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\times \displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{(h+1)-1}{h\times \left ( \sqrt{h+1}+1 \right )}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt{h+1}+1 \right )}\\ &=\displaystyle \frac{1}{\sqrt{0+1}+1}\\ &=\color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textbf{(Mat Das SIMAK UI 2013)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow 5}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}+\sqrt{2}&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 5-2\sqrt{6}\quad &\textrm{c}.\quad 2\sqrt{6}\quad &\textrm{e}.\quad \color{red}5+2\sqrt{6}\end{array}\\\\ &\textrm{Jawab}:\\ & \begin{aligned}\underset{x\rightarrow 5}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}\\ &=\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{3+2+2\sqrt{6}}{3-2}\\ &=\color{red}5+2\sqrt{6} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 9.&\textbf{(Mat IPA SBMPTN 2014)}\\\\ &\textrm{Jika}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4\\ & \textrm{dan}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3,\\\\ &\textrm{maka nilai}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x).g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{14}&&\textrm{d}.\quad \displaystyle \frac{4}{14}\\\\ \textrm{b}.\quad \color{red}\displaystyle \frac{2}{14}\quad &\textrm{c}.\quad \displaystyle \frac{3}{14}\quad &\textrm{e}.\quad \displaystyle \frac{5}{14}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: ,\\ &\begin{array}{lll}\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4&&\\ &&\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3&+&\\ &&\\\hline &&\\ 2\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x)\qquad\qquad\: \: =1&&\\ &&\\ \qquad\qquad\: \: \: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x)=\displaystyle \frac{1}{2},&&\\\end{array}\\ &\textrm{sehingga}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(g)=\displaystyle \frac{2}{7}\\ &&\\ &\textrm{maka},\\ &\underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x).g(x)=\displaystyle \frac{1}{2}\times \frac{2}{7}=\color{red}\displaystyle \frac{2}{14} \end{aligned} \end{array}$

Contoh Soal 1 Limit Fungsi Aljabar

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

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

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

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

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