Contoh Soal 4 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{l}\\ 16.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{4}{9}}\\ \\ \text{c}.& {\displaystyle \frac{1}{2}}\\ \\ \text{d}.& 1\\ \\ \text{e}.& {\displaystyle \frac{3}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}\times {\displaystyle \frac{\left(\sqrt{{\displaystyle \frac{1}{x^2}}}\right)}{\left(\sqrt{{\displaystyle \frac{1}{x^2}}}\right)}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^2}}}{\sqrt{9-\frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}}\\ & ={\displaystyle \frac{2-1}{3}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x^4+2x^3-5x+2021}{2x^3-4x^2+2020}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{4}{9}}\\ \\ \text{b}.{\displaystyle \frac{3}{2}}\\ \\ \text{c}.{\displaystyle 0}& \\ \\ \text{d}.{\displaystyle 1}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x^4+2x^3-5x+2021}{2x^3-4x^2+2020}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3x^4}{x^4}+\frac{2x^3}{x^4}-\frac{5x}{x^4}+\frac{2021}{x^4}}}{{\displaystyle \frac{2x^3}{x^4}-\frac{4x^2}{x^4}+\frac{2020}{x^4}}}}}\\ & ={\displaystyle \frac{3+{\displaystyle \frac{2}{x}-\frac{5}{x^2}+\frac{2021}{x^4}}}{{\displaystyle \frac{4}{x}-\frac{4}{x^2}+\frac{2020}{x^4}}}}\\ & ={\displaystyle \frac{3+0-0+0}{0-0+0}}\\ & =\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{x^2+3x+4}{3x^2+2x+3}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{4}{3}}\\ \\ \text{b}.{\displaystyle \frac{1}{3}}\\ \\ \text{c}.{\displaystyle 0}\\ \\ \text{d}.{\displaystyle 3}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{x^2+3x+4}{3x^2+2x+3}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{x^2}{x^2}+\frac{3x}{x^2}+\frac{4}{x^2}}}{{\displaystyle \frac{3x^2}{x^2}+\frac{2x}{x^2}+\frac{3}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{1+{\displaystyle \frac{3}{x}+\frac{4}{x^2}}}{3+{\displaystyle \frac{2}{x}+\frac{3}{x^2}}}}\\ & ={\displaystyle \frac{1+0+0}{3+0+0}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x}{9x^2+x+1}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 3}\\ \\ \text{b}.{\displaystyle 1}\\ \\ \text{c}.{\displaystyle \frac{1}{3}}\\ \\ \text{d}.{\displaystyle 0}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x}{9x^2+x+1}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3x}{x^2}}}{{\displaystyle \frac{9x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3}{x}}}{{\displaystyle \frac{9x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}}\\ & ={\displaystyle \frac{0}{9+0+0}}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2-2x-8}-\sqrt{x^2+2x+1})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -2}& & \\ \text{b}.{\displaystyle -1}\\ \text{c}.{\displaystyle -\frac{1}{2}}\\ \text{d}.{\displaystyle 0}\\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2-2x-8}-\sqrt{x^2+2x+1})\\ & =\mathrm{\infty }-\mathrm{\infty }=\mathbf{\text{tidak diperbolehkan}}\\ & \text{Selanjutnya gunakan formula}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})={\displaystyle \frac{b-p}{2\sqrt{a}},\text{maka}}\\ & ={\displaystyle \frac{-2-2}{2\sqrt{1}}}\\ & =\frac{-4}{2}\\ & =-2\end{array}\end{array}$