Contoh Soal 5 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{l}\\ 21.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}(x-\sqrt{x^2-10x})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -10}\\ \text{b}.{\displaystyle -5}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 5}\\ \text{e}.{\displaystyle 10}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(x-\sqrt{x^2-10x})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2}-\sqrt{x^2-10x})\\ & \text{Selanjutnya gunakan formula}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})\\ & ={\displaystyle \frac{b-p}{2\sqrt{a}},\text{maka}}\\ & ={\displaystyle \frac{0-(-10)}{2\sqrt{1}}}\\ & =\frac{10}{2}\\ & =5\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}5\tan {\displaystyle \frac{1}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \mathrm{\infty }}\\ \text{b}.& {\displaystyle 5}\\ \text{c}.& {\displaystyle \sqrt{3}}\\ \text{d}.& 1\\ \text{e}.& {\displaystyle 0}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}5\tan {\displaystyle \frac{1}{x}}& =\underset{u\to 0}{\text{Lim}}5\tan {\displaystyle u}\\ & =5\tan 0\\ & =5.\mathrm{\infty }\\ & =\mathrm{\infty }\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}15x\tan {\displaystyle \frac{4}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0}\\ \text{b}.& {\displaystyle \frac{1}{4}}\\ \text{c}.& {\displaystyle 4}\\ \text{d}.& {\displaystyle \frac{11}{4}}\\ \text{e}.& {\displaystyle 60}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}15x\tan {\displaystyle \frac{4}{x}}& =\cdots \\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}\text{maka}x={\displaystyle \frac{1}{u}}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{15}{u}.\tan 4u}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{15\tan 4u}{u}}\\ & =15\times 4\\ & =60\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{x}^2{\sin }^2\left({\displaystyle \frac{ab}{x}}\right)\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle ab}\\ \text{b}.& {\displaystyle a^{2}b}\\ \text{c}.& {\displaystyle a{b}^2}\\ \text{d}.& {\displaystyle (ab{)}^2}\\ \text{e}.& {\displaystyle \frac{1}{(ab{)}^2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{x}^2\sin {\displaystyle \frac{ab}{x}}& =\underset{u\to 0}{\text{Lim}}{\left({\displaystyle \frac{1}{u}}\right)}^2{\sin }^2{\displaystyle abu}\\ & =\underset{u\to 0}{\text{Lim}}\left({\displaystyle \frac{{\sin }^{2}abu}{u^2}}\right)\\ & =(ab{)}^2\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{{\displaystyle \frac{x}{100}}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\mathrm{\infty }}\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{{\displaystyle \frac{x}{100}}}}& =100\times \underset{0}{\underbrace{\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{x}}}}\\ & =100\times 0\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 26.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{18}{x\sin {\displaystyle \frac{3}{x}}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -54}\\ \text{b}.& {\displaystyle -6}\\ \text{c}.& {\displaystyle {\displaystyle \frac{1}{6}}}\\ \text{d}.& {\displaystyle 6}\\ \text{e}.& {\displaystyle 54}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{18}{x\sin {\displaystyle \frac{3}{x}}}}& =\cdots \\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}\text{maka}x={\displaystyle \frac{1}{u}}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{18}{{\displaystyle \frac{1}{u}\sin 3u}}}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{18u}{\sin 3u}}\\ & ={\displaystyle \frac{18}{3}}\\ & =6\end{array}\end{array}$.

$\begin{array}{l}\\ 27.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{4x\sin {\displaystyle \frac{2}{x}}}{2}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -4}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& {\displaystyle {\displaystyle \frac{1}{2}}}\\ \text{d}.& {\displaystyle 2}\\ \text{e}.& {\displaystyle 4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{4x\sin {\displaystyle \frac{2}{x}}}{2}}& =\cdots \\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}\text{maka}x={\displaystyle \frac{1}{u}}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{4}{u}.\sin 2u}}{2}}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{4\sin 2u}{2u}}\\ & ={\displaystyle \frac{4\times 2}{2}}\\ & =4\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Nilai dari}\\ & \underset{x\to -\mathrm{\infty }}{\text{Lim}}{\displaystyle x\cos \frac{1}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\mathrm{\infty }}\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\underset{x\to -\mathrm{\infty }}{\text{Lim}}{\displaystyle x\cos \frac{1}{x}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (-x)\cos \frac{1}{(-x)}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (-x)\cos \frac{1}{(x)}}\\ & =-\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (x)\cos \frac{1}{(x)}}\\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =-\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{1}{u}\cos u}\\ & =-\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{\cos u}{u}}\\ & =-{\displaystyle \frac{1}{0}}\\ & =-\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 29.& \text{Asimtot tegak dari fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x=2\text{dan}x=4\\ \text{b}.& x=2\text{dan}x=3\\ \text{c}.& x=3\text{dan}x=4\\ \text{d}.& x=3\text{saja}\\ \text{e}.& x=2\text{saja}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Asimtot tegak fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}}\\ & \text{terjadi saat penyebut}=0.\\ & \text{Sehingga}{x}^2-5x+6=0\\ & \iff (x-2)(x-3)=0,\text{maka}\\ & x=2\text{atau}x=3\\ & \therefore \text{asimtot tegak fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}}\\ & \text{adalah}x=2\text{dan}x=3\end{array}\end{array}$

$\begin{array}{l}\\ 30.& \text{Asimtot datar dari fungsi}\\ & g(x)={\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& y=-3\\ \text{b}.& y=-1\\ \text{c}.& {\displaystyle \frac{1}{3}}\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Asim}& \text{tot datar dari fungsi}\\ g(x)& ={\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\text{untuk}}\\ g(x)& ={\displaystyle \frac{(6x^2-8x+2)}{(-2x^2+5x-2)}\text{terjadi saat}}\\ y& ={\displaystyle \frac{6}{-2}=-3}\\ & \mathbf{\text{atau dapat juga dicari}}\mathbf{\text{dengan}}\\ y& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(6x^2-8x+2)}{(-2x^2+5x-2)}\times {\displaystyle \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{6-{\displaystyle \frac{8}{x}+\frac{2}{x^2}}}{-2+{\displaystyle \frac{5}{x}-\frac{2}{x^2}}}}\\ & ={\displaystyle \frac{6-0+0}{-2+0-0}}\\ & ={\displaystyle \frac{6}{-2}}\\ & =-3\end{array}\end{array}$