Contoh Soal 2 Limit Fungsi Aljabar

$\begin{array}{l}\\ 6.& \text{Diketahui bahwa}f(x)=x^2-2,\\ & \text{maka nilai}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x^2-2}& & \text{d}.{\displaystyle x}\\ \\ \text{b}.{\displaystyle x^2}& \text{c}.{\displaystyle 2x}& \text{e}.{\displaystyle 2x-2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}f(x)=x^2-2,\\ \text{maka nila}& \text{i untuk}\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{((x+h{)}^2-2)-(x^2-2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{x^2+2xh+h^2-2-x^2+2}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{2xh+h^2}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle (2x+h)}\\ & =2x\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Diketahui}f(x)=\sqrt{x-1},\\ & \text{maka nilai}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}}& & \text{d}.{\displaystyle 1}\\ \\ \text{b}.-{\displaystyle \frac{1}{2}}& \text{c}.{\displaystyle 0}& \text{e}.{\displaystyle -1}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}f(x)=\sqrt{x-1},\\ \text{maka}& \text{nilai untuk}\\ \underset{h\to 0}{\text{Lim}}& {\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{\sqrt{h+1}-1}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle {\displaystyle \frac{\sqrt{h+1}-1}{h}\times {\displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(h+1)-1}{h\times (\sqrt{h+1}+1)}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{1}{(\sqrt{h+1}+1)}}\\ & ={\displaystyle \frac{1}{\sqrt{0+1}+1}}\\ & ={\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \mathbf{\text{(Mat Das SIMAK UI 2013)}}\\ & \text{Nilai}\underset{x\to 5}{\text{Lim}}{\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....}\\ & \begin{array}{l}\\ \text{a}.\sqrt{3}+\sqrt{2}& & \text{d}.5\\ \text{b}.5-2\sqrt{6}& \text{c}.2\sqrt{6}& \text{e}.5+2\sqrt{6}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\underset{x\to 5}{\text{Lim}}& {\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}}\\ & ={\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}}\\ & ={\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}}\\ & ={\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}}\\ & ={\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times {\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}}}\\ & ={\displaystyle \frac{3+2+2\sqrt{6}}{3-2}}\\ & =5+2\sqrt{6}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \mathbf{\text{(Mat IPA SBMPTN 2014)}}\\ \\ & \text{Jika}\underset{x\to a}{\text{Lim}}{\displaystyle (f(x)+\frac{1}{g(x)})=4}\\ & \text{dan}\underset{x\to a}{\text{Lim}}{\displaystyle (f(x)-\frac{1}{g(x)})=-3,}\\ \\ & \text{maka nilai}\underset{x\to a}{\text{Lim}}{\displaystyle f(x).g(x)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{14}}& & \text{d}.{\displaystyle \frac{4}{14}}\\ \\ \text{b}.{\displaystyle \frac{2}{14}}& \text{c}.{\displaystyle \frac{3}{14}}& \text{e}.{\displaystyle \frac{5}{14}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa},\\ & \begin{array}{l}\\ \underset{x\to a}{\text{Lim}}{\displaystyle (f(x)+\frac{1}{g(x)})=4}& & \\ & & \\ \underset{x\to a}{\text{Lim}}{\displaystyle (f(x)-\frac{1}{g(x)})=-3}& +& \\ & & \\ & & \\ 2\underset{x\to a}{\text{Lim}}{\displaystyle f(x)=1}& & \\ & & \\ \underset{x\to a}{\text{Lim}}f(x)={\displaystyle \frac{1}{2},}& & \end{array}\\ & \text{sehingga}\underset{x\to a}{\text{Lim}}f(g)={\displaystyle \frac{2}{7}}\\ & & \\ & \text{maka},\\ & \underset{x\to a}{\text{Lim}}f(x).g(x)={\displaystyle \frac{1}{2}\times \frac{2}{7}={\displaystyle \frac{2}{14}}}\end{array}\end{array}$

Contoh Soal 1 Limit Fungsi Aljabar

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

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

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

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

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