Contoh Soal 1 Limit Fungsi Aljabar

$\begin{array}{l}\\ 1& \text{Nilai}\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)=....\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle \frac{1}{2}}& & \text{d}.{\displaystyle \frac{1}{4}}\\ \\ \text{b}.-{\displaystyle \frac{1}{4}}& \text{c}.0& \text{e}.{\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)& =\left({\displaystyle \frac{6-2}{2^2-4}-\frac{1}{2-2}}\right)\\ & =\left({\displaystyle \frac{4}{0}-\frac{1}{0}}\right)=\mathrm{\infty }-\mathrm{\infty }\\ & \text{hal ini tidak diperkenankan}\\ \text{Sehingga},& \\ \underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{1}{x-2}}\right)& =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{(x+2)}{(x-2)(x+2)}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{6-x}{x^2-4}-\frac{x+2}{x^2-4}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{4-2x}{x^2-4}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{-2(x-2)}{(x+2)(x-2)}}\right)\\ & =\underset{x\to 2}{\text{Lim}}\left({\displaystyle \frac{-2}{x+2}}\right)\\ & ={\displaystyle -\frac{2}{(2+2)}}\\ & =-{\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nilai}\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}=....}\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle 2}& & \text{d}.{\displaystyle \frac{1}{2}}\\ \\ \text{b}.-{\displaystyle \frac{1}{2}}& \text{c}.0& \text{e}.{\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}}& =\left({\displaystyle \frac{4-4}{2^4-4}}\right)\\ & ={\displaystyle \frac{0}{0}}\\ & \text{hal ini juga tidak diperkenankan}\\ \text{Sehingga},& \\ \underset{x\to 4}{\text{Lim}}{\displaystyle \frac{x-4}{2\sqrt{x}-x}}& =\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{(\sqrt{x}+2)(\sqrt{x}-2)}{\sqrt{x}(2-\sqrt{x})}}\\ & =\underset{x\to 4}{\text{Lim}}{\displaystyle \frac{(\sqrt{x}+2)(\sqrt{x}-2)}{-\sqrt{x}(\sqrt{x}-2)}}\\ & =\underset{x\to 4}{\text{Lim}}-{\displaystyle \frac{(\sqrt{x}+2)}{\sqrt{x}}}\\ & =-{\displaystyle \frac{\sqrt{4}+2}{\sqrt{4}}}\\ & =-{\displaystyle \frac{2+2}{2}}\\ & =-2\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Nilai}\underset{x\to 1}{\text{Lim}}{\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}}& & \text{d}.{\displaystyle 2}\\ \\ \text{b}.{\displaystyle \frac{1}{2}}& \text{c}.1& \text{e}.{\displaystyle 4}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 1}{\text{Lim}}& {\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}}\\ & ={\displaystyle \frac{(2-3+1)(1-1)}{{(1-1)}^2}}\\ & ={\displaystyle \frac{0\times 0}{0^2}}\\ & ={\displaystyle \frac{0}{0}}\\ & \text{hal ini juga tidak diperkenankan}\\ \text{Sehi}& \text{ngga},\\ \underset{x\to 1}{\text{Lim}}& {\displaystyle \frac{(2x-3\sqrt{x}+1)(\sqrt{x}-1)}{{(\sqrt{x}-1)}^2}}\\ & =\underset{x\to 1}{\text{Lim}}{\displaystyle \frac{((2\sqrt{x}-1)\times (\sqrt{x}-1))(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}-1)}}\\ & =\underset{x\to 1}{\text{Lim}}(2\sqrt{x}-1)\\ & =2.1-1\\ & =2-1\\ & =1\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Nilai}\underset{x\to 3}{\text{Lim}}{\displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}=....}\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle \frac{1}{14}\sqrt{7}}& & \text{d}.{\displaystyle \frac{1}{7}\sqrt{7}}\\ \\ \text{b}.-{\displaystyle \frac{1}{7}\sqrt{7}}& \text{c}.0& \text{e}.{\displaystyle \frac{1}{14}\sqrt{7}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 3}{\text{Lim}}& {\displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}}\\ & =\underset{x\to 3}{\text{Lim}}{\displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\times {\displaystyle \frac{\sqrt{x+4}+\sqrt{2x+1}}{\sqrt{x+4}+\sqrt{2x+1}}}}\\ & =\underset{x\to 3}{\text{Lim}}{\displaystyle \frac{(x+4)-(2x+1)}{(x-3)(\sqrt{x+4}+\sqrt{2x+1})}}\\ & =\underset{x\to 3}{\text{Lim}}-{\displaystyle \frac{-x+3}{(x-3)(\sqrt{x+4}+\sqrt{2x+1})}}\\ & =\underset{x\to 3}{\text{Lim}}{\displaystyle \frac{-1}{(\sqrt{x+4}+\sqrt{2x+1})}}\\ & =-{\displaystyle \frac{1}{(\sqrt{7}+\sqrt{7})}}\\ & =-{\displaystyle \frac{1}{2\sqrt{7}}}\\ & =-{\displaystyle \frac{1}{2\sqrt{7}}\times {\displaystyle \frac{\sqrt{7}}{\sqrt{7}}}}\\ & =-{\displaystyle \frac{1}{14}\sqrt{7}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Jika}\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6,\text{maka nilai}a\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 2}& & \text{d}.{\displaystyle -2}\\ \\ \text{b}.{\displaystyle 1}& \text{c}.-1& \text{e}.{\displaystyle -3}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 2}{\text{Lim}}& {\displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6}\\ & \text{dengan bantuan limit kanan }\\ & \text{yaitu}x=2+h\Rightarrow h\to 0\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{a(2+h)-2a}{\sqrt{2(2+h)}-(2+h)}=6}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2a+ah-2a}{\sqrt{4+2h}-(2+h)}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah}{\sqrt{4+2h}-(2+h)}\times {\displaystyle \frac{(\sqrt{4+2h}+(2+h))}{(\sqrt{4+2h}+(2+h))}}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah\times (\sqrt{4+2h}+(2+h))}{4+2h-(4+4h+h^2)}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah\times (\sqrt{4+2h}+(2+h))}{-2h-h^2}}\\ 6& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a\times (\sqrt{4+2h}+(2+h))}{-2-h}}\\ 6& ={\displaystyle \frac{a\times (\sqrt{4+0}+(2+0))}{-2-0}}\\ 6& ={\displaystyle \frac{a(\sqrt{4}+2)}{-2}}\\ {\displaystyle \frac{a(4)}{-2}}& =6\\ a(-2)& =6\\ a& =-3\end{array}\end{array}$