Limit Fungsi Aljabar untuk x Mendekati Tak Hingga

A. Bentuk $\lim\limits_{x \to \infty}\frac{f(x)}{g(x)}$

Konsep:
Jika kita lakukan substitusi langsung pada bentuk limit tersebut maka akan diperoleh bentuk $\frac{\infty }{\infty }$. Penyelesaian bentuk $\lim\limits_{x \to \infty}\frac{f(x)}{g(x)}$ dilakukan dengan cara membagi $f(x)$ dan $g(x)$ dengan ${{x}^{n}}$ dimana $n$ adalah pangkat tertinggi dari $f(x)$ atau $g(x)$.
Cara Praktis:
$\lim\limits_{x \to \infty}\frac{ax^n+bx^{n-1}+\cdots +c}{px^m+qx^{m-1}+\cdots +r}=L$, maka:

• jika $n < m$ maka $L=0$
• jika $n=m$ maka $L=\frac{a}{p}$
• jika $n > m$ maka:
1. $L=\infty $ jika $\frac{a}{p} > 0$
2. $L=-\infty $ jika $\frac{a}{p} < 0$

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Contoh 1.
Tentukan nilai $\lim\limits_{x \to \infty}\frac{2x^4-3}{x^2+4x-5}$.
Penyelesaian:
Konsep:
$\begin{align}\lim\limits_{x \to \infty}\frac{2x^4-3}{x^2+4x-5} &= \lim\limits_{x \to \infty}\frac{\frac{2x^4}{x^4}-\frac{3}{x^4}}{\frac{x^2}{x^4}+\frac{4x}{x^4}-\frac{5}{x^4}} \\ &= \lim\limits_{x \to \infty}\frac{2-\frac{3}{x^4}}{\frac{1}{x^2}+\frac{4}{x^3}-\frac{5}{x^4}} \\ &= \frac{2-0}{0+0-0} \\ &= \frac{2}{0} \\ &= \infty \end{align}$
Cara Praktis:
$\lim\limits_{x \to \infty}\frac{2x^4-3}{x^2+4x-5}$ = ...
$n=4$, $m=2$ dan $n > m$ maka $\lim\limits_{x \to \infty}\frac{2x^4-3}{x^2+4x-5}=\infty $.

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Contoh 2.
Tentukan nilai $\lim\limits_{x \to \infty}\frac{4x^2+2x-3}{7x^2+x+5}$.
Penyelesaian:
Konsep:
$\begin{align}\lim\limits_{x \to \infty}\frac{4x^2+2x-3}{7x^2+x+5} &= \lim\limits_{x \to \infty}\frac{\frac{4x^2}{x^2}+\frac{2x}{x^2}-\frac{3}{x^2}}{\frac{7x^2}{x^2}+\frac{x}{x^2}+\frac{5}{x^2}} \\ &= \lim\limits_{x \to \infty}\frac{4+\frac{2}{x}-\frac{3}{x^2}}{7+\frac{1}{x}+\frac{5}{x^2}} \\ &= \frac{4+0-0}{7+0+0} \\ &= \frac{4}{7} \end{align}$
Cara Praktis:
$\lim\limits_{x \to \infty}\frac{4x^2+2x-3}{7x^2+x+5}$ = ...
$n=2$, $m=2$ dan $n=m$ maka:
$\lim\limits_{x \to \infty}\frac{4x^2+2x-3}{7x^2+x+5}=\frac{4}{7}$

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Contoh 3.
Tentukan nilai $\lim\limits_{x \to \infty}\frac{2x+5}{3x^3+2x^2+1}$.
Penyelesaian:
Konsep:
$\begin{align}\lim\limits_{x \to \infty}\frac{2x+5}{3x^3+2x^2+1} &= \lim\limits_{x \to \infty}\frac{\frac{2x}{x^3}+\frac{5}{x^3}}{\frac{3x^3}{x^3}+\frac{2x^2}{x^3}+\frac{1}{x^3}} \\ &= \lim\limits_{x \to \infty}\frac{\frac{2}{x^2}+\frac{5}{x^3}}{3+\frac{2}{x}+\frac{1}{x^3}} \\ &= \frac{0+0}{3+0+0} \\ &= \frac{0}{3} \\ &= 0 \end{align}$
Cara Praktis:
$\lim\limits_{x \to \infty}\frac{2x+5}{3x^3+2x^2+1}$ = ...
$n=1$, $m=3$, dan $n < m$ maka $\lim\limits_{x \to \infty}\frac{2x+5}{3x^3+2x^2+1}=0$.

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Contoh 4.
Tentukan nilai $\lim\limits_{x \to \infty}\frac{3^x+3^{x+2}}{3^x+3^{x-1}}$.
Penyelesaian:
$\begin{align}\lim\limits_{x \to \infty}\frac{3^x+3^{x+2}}{3^x+3^{x-1}} &= \lim\limits_{x \to \infty}\frac{\frac{3^x}{3^x}+\frac{3^{x+2}}{3^x}}{\frac{3^x}{3^x}+\frac{3^{x-1}}{3^x}} \\ &= \lim\limits_{x \to \infty}\frac{1+3^2}{1+3^{-1}} \\ &= \frac{1+9}{1+\frac{1}{9}} \\ &= 10\times \frac{9}{10} \\ = 9 \end{align}$

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B. Bentuk $\lim\limits_{x \to \infty}(\sqrt{f(x)}-\sqrt{g(x)})$

Konsep:
Bentuk $\lim\limits_{x \to \infty}(\sqrt{f(x)}-\sqrt{g(x)})$ dapat diselesaikan dengan cara mengalikan dengan bentuk sekawannya, yaitu $\frac{\sqrt{f(x)}+\sqrt{g(x)}}{\sqrt{f(x)}+\sqrt{g(x)}}$, kemudian membaginya dengan $x$ pangkat tertinggi dari pembilang atau penyebutnya.
Cara Praktis: $\lim\limits_{x \to \infty}(\sqrt{ax+b}-\sqrt{px+q})=L$ maka:

• Jika $a < p$ maka $L=-\infty $
• Jika $a=p$ maka $L=0$
• Jika $a > p$ maka $L=\infty $

$\lim\limits_{x \to \infty}(\sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r})$ = L maka:

• Jika $a < p$ maka $L=-\infty $
• Jika $a=p$ maka $L=\frac{b-q}{2\sqrt{a}}$
• Jika $a > p$ maka $L=\infty $

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Contoh 1.
Tentukan nilai $\lim\limits_{x \to \infty}(\sqrt{4x-3}-\sqrt{5x-2})$.
Penyelesaian:
$\lim\limits_{x \to \infty}(\sqrt{4x-3}-\sqrt{5x-2})$
= $\lim\limits_{x \to \infty}(\sqrt{4x-3}-\sqrt{5x-2})\cdot \frac{(\sqrt{4x-3}+\sqrt{5x-2})}{(\sqrt{4x-3}+\sqrt{5x-2})}$
= $\lim\limits_{x \to \infty}\frac{4x-3-(5x-2)}{\sqrt{4x-3}+\sqrt{5x-2}}$
= $\lim\limits_{x \to \infty}\frac{-x-1}{\sqrt{4x-3}+\sqrt{5x-2}}$
= $\lim\limits_{x \to \infty}\frac{\frac{-x}{x}-\frac{1}{x}}{\sqrt{\frac{4x}{x^2}-\frac{3}{x^2}}+\sqrt{\frac{5x}{x^2}-\frac{2}{x^2}}}$
= $\lim\limits_{x \to \infty}\frac{-1-\frac{1}{x}}{\sqrt{\frac{4}{x}-\frac{3}{x^2}}+\sqrt{\frac{5}{x}-\frac{2}{x^2}}}$
= $\frac{-1-0}{\sqrt{0-0}+\sqrt{0-0}}$
= $\frac{-1}{0}$
= $-\infty $
Cara Praktis:
$\lim\limits_{x \to \infty}(\sqrt{4x-3}-\sqrt{5x-2})$
$a=4$, $p=5$, dan $a < p$ maka:
$\lim\limits_{x \to \infty}(\sqrt{4x-3}-\sqrt{5x-2})=-\infty $

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Contoh 2.
Tentukan nilai $\lim\limits_{x \to \infty}(\sqrt{9x+5}-\sqrt{4x+1})$.
Penyelesaian:
Konsep:
$\lim\limits_{x \to \infty}(\sqrt{9x+5}-\sqrt{4x+1})$
= $\lim\limits_{x \to \infty}(\sqrt{9x+5}-\sqrt{4x+1})\cdot \frac{(\sqrt{9x+5}+\sqrt{4x+1})}{(\sqrt{9x+5}+\sqrt{4x+1})}$
= $\lim\limits_{x \to \infty}\frac{(9x+5)-(4x+1)}{\sqrt{9x+5}+\sqrt{4x+1}}$
= $\lim\limits_{x \to \infty}\frac{5x+4}{\sqrt{9x+5}+\sqrt{4x+1}}$
= $\lim\limits_{x \to \infty}\frac{\frac{5x}{x}+\frac{4}{x}}{\sqrt{\frac{9x}{x^2}+\frac{5}{x^2}}+\sqrt{\frac{4x}{x^2}+\frac{1}{x^2}}}$
= $\lim\limits_{x \to \infty}\frac{5+\frac{4}{x}}{\sqrt{\frac{9}{x}+\frac{5}{x^2}}+\sqrt{\frac{4}{x}+\frac{1}{x^2}}}$
= $\frac{5+0}{\sqrt{0+0}+\sqrt{0+0}}$
= $\frac{5}{0}$
= $\infty $
Cara Praktis:
$\lim\limits_{x \to \infty}(\sqrt{9x+5}-\sqrt{4x+1})$
$a=9$, $p=4$, dan $a > p$ maka:
$\lim\limits_{x \to \infty}(\sqrt{9x+5}-\sqrt{4x+1})=\infty $

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Contoh 3.
Tentukan nilai $\lim\limits_{x \to \infty}(\sqrt{7x-8}-\sqrt{7x+4})$.
Penyelesaian:
$\lim\limits_{x \to \infty}(\sqrt{7x-8}-\sqrt{7x+4})$
= $\lim\limits_{x \to \infty}(\sqrt{7x-8}-\sqrt{7x+4})\cdot \frac{(\sqrt{7x-8}+\sqrt{7x+4})}{(\sqrt{7x-8}+\sqrt{7x+4})}$
= $\lim\limits_{x \to \infty}\frac{(7x-8)-(7x+4)}{\sqrt{7x-8}+\sqrt{7x+4}}$
= $\lim\limits_{x \to \infty}\frac{-12}{\sqrt{7x-8}+\sqrt{7x+4}}$
= $\lim\limits_{x \to \infty}\frac{\frac{-12}{\sqrt{x}}}{\frac{\sqrt{7x-8}+\sqrt{7x+4}}{\sqrt{x}}}$
= $\lim\limits_{x \to \infty}\frac{\frac{-12}{\sqrt{x}}}{\sqrt{\frac{7x}{x}-\frac{8}{x}}+\sqrt{\frac{7x}{x}+\frac{4}{x}}}$
= $\lim\limits_{x \to \infty}\frac{\frac{-12}{\sqrt{x}}}{\sqrt{7-\frac{8}{x}}+\sqrt{7+\frac{4}{x}}}$
= $\frac{0}{\sqrt{7-0}+\sqrt{7+0}}$
= $\frac{0}{2\sqrt{7}}$
= 0
Cara Praktis:
$\lim\limits_{x \to \infty}(\sqrt{7x-8}-\sqrt{7x+4})$
$a=7$, $p=7$ dan $a=p$ maka:
$\lim\limits_{x \to \infty}(\sqrt{7x-8}-\sqrt{7x+4})=0$

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Contoh 4.
Tentukan nilai $\lim\limits_{x \to \infty}(\sqrt{2x^2+3x-1}-\sqrt{x^2-x+2})$.
Penyelesaian:
Konsep:
$\lim\limits_{x \to \infty}(\sqrt{2x^2+3x-1}-\sqrt{x^2-x+2})$
= $\lim\limits_{x \to \infty}(\sqrt{2x^2+3x-1}-\sqrt{x^2-x+2})$ $\times$ $\frac{(\sqrt{2x^2+3x-1}+\sqrt{x^2-x+2})}{(\sqrt{2x^2+3x-1}+\sqrt{x^2-x+2})}$
= $\lim\limits_{x \to \infty}\frac{(2x^2+3x-1)-(x^2-x+2)}{\sqrt{2x^2+3x-1}+\sqrt{x^2-x+2}}$
= $\lim\limits_{x \to \infty}\frac{2x^2+3x-1-x^2+x-2}{\sqrt{2x^2+3x-1}+\sqrt{x^2-x+2}}$
= $\lim\limits_{x \to \infty}\frac{x^2+4x-3}{\sqrt{2x^2+3x-1}+\sqrt{x^2-x+2}}$
= $\lim\limits_{x \to \infty}\frac{\frac{x^2+4x-3}{x^2}}{\frac{\sqrt{2x^2+3x-1}+\sqrt{x^2-x+2}}{x^2}}$
= $\lim\limits_{x \to \infty}\frac{\frac{x^2}{x^2}+\frac{4x}{x^2}-\frac{3}{x^2}}{\sqrt{\frac{2x^2}{x^4}+\frac{3x}{x^4}-\frac{1}{x^4}}+\sqrt{\frac{x^2}{x^4}-\frac{x}{x^4}+\frac{2}{x^4}}}$
= $\lim\limits_{x \to \infty}\frac{1+\frac{4}{x}-\frac{3}{x^2}}{\sqrt{\frac{2}{x^2}+\frac{3}{x^3}-\frac{1}{x^4}}+\sqrt{\frac{1}{x^2}-\frac{1}{x^3}+\frac{2}{x^4}}}$
= $\lim\limits_{x \to \infty}\frac{1+\frac{4}{\infty }-\frac{3}{\infty }}{\sqrt{\frac{2}{\infty }+\frac{3}{\infty }-\frac{1}{\infty }}+\sqrt{\frac{1}{\infty }-\frac{1}{\infty }+\frac{2}{\infty }}}$
= $\frac{1+0-0}{\sqrt{0+0-0}+\sqrt{0-0+0}}$
= $\frac{1}{0}$
= $\infty $
Cara Praktis:
$\lim\limits_{x \to \infty}(\sqrt{2x^2+3x-1}-\sqrt{x^2-x+2})$
$a=2$, $p=1$, dan $a > p$ maka:
$\lim\limits_{x \to \infty}(\sqrt{2x^2+3x-1}-\sqrt{x^2-x+2})=\infty $

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Contoh 5.
Tentukan nilai $\lim\limits_{x \to \infty}(2x+3-\sqrt{4x^2-5x+3})$.
Penyelesaian:
$\lim\limits_{x \to \infty}(2x+3-\sqrt{4x^2-5x+3})$
= $\lim\limits_{x \to \infty}(\sqrt{(2x+3)^2}-\sqrt{4x^2-5x+3})$
= $\lim\limits_{x \to \infty}(\sqrt{4x^2+12x+9}-\sqrt{4x^2-5x+3})$
= $\lim\limits_{x \to \infty}(\sqrt{4x^2+12x+9}-\sqrt{4x^2-5x+3})$ $\times$ $\frac{(\sqrt{4x^2+12x+9}+\sqrt{4x^2-5x+3})}{(\sqrt{4x^2+12x+9}+\sqrt{4x^2-5x+3})}$
= $\lim\limits_{x \to \infty}\frac{(4x^2+12x+9)-(4x^2-5x+3)}{\sqrt{4x^2+12x+9}+\sqrt{4x^2-5x+3}}$
= $\lim\limits_{x \to \infty}\frac{17x+6}{\sqrt{4x^2+12x+9}+\sqrt{4x^2-5x+3}}$
= $\lim\limits_{x \to \infty}\frac{\frac{17x}{x}+\frac{6}{x}}{\sqrt{\frac{4x^2}{x^2}+\frac{12x}{x^2}+\frac{9}{x^2}}+\sqrt{\frac{4x^2}{x^2}-\frac{5x}{x^2}+\frac{3}{x^2}}}$
= $\lim\limits_{x \to \infty}\frac{17+\frac{6}{x}}{\sqrt{4+\frac{12}{x}+\frac{9}{x^2}}+\sqrt{4-\frac{5}{x}+\frac{3}{x^2}}}$
= $\frac{17+0}{\sqrt{4+0+0}+\sqrt{4-0+0}}$
= $\frac{17}{2\sqrt{4}}$
= $\frac{17}{4}$
Cara Praktis:
$\lim\limits_{x \to \infty}(2x+3-\sqrt{4x^2-5x+3})$
= $\lim\limits_{x \to \infty}(\sqrt{(2x+3)^2}-\sqrt{4x^2-5x+3})$
= $\lim\limits_{x \to \infty}(\sqrt{4x^2+12x+9}-\sqrt{4x^2-5x+3})$
$a=4$, $b=12$, $p=4$, $q=-5$, dan $a=p$ maka:
= $\frac{b-q}{2\sqrt{a}}$
= $\frac{12-(-5)}{2\sqrt{4}}$
= $\frac{17}{4}$

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Contoh 6.
Tentukan nilai $\lim\limits_{x \to \infty}(\sqrt{3x^2-2x+7}-\sqrt{4x^2+3x+5})$.
Penyelesaian:
$\lim\limits_{x \to \infty}(\sqrt{3x^2-2x+7}-\sqrt{4x^2+3x+5})$
= $\lim\limits_{x \to \infty}(\sqrt{3x^2-2x+7}-\sqrt{4x^2+3x+5})$ $\times$ $\frac{(\sqrt{3x^2-2x+7}+\sqrt{4x^2+3x+5})}{(\sqrt{3x^2-2x+7}+\sqrt{4x^2+3x+5})}$
= $\lim\limits_{x \to \infty}\frac{(3x^2-2x+7)-(4x^2+3x+5)}{\sqrt{3x^2-2x+7}+\sqrt{4x^2+3x+5}}$
= $\lim\limits_{x \to \infty}\frac{-x^2-5x+2}{\sqrt{3x^2-2x+7}+\sqrt{4x^2+3x+5}}$
= $\lim\limits_{x \to \infty}\frac{\frac{-x^2}{x^2}-\frac{5x}{x^2}+\frac{2}{x^2}}{\sqrt{\frac{3x^2}{x^4}-\frac{2x}{x^4}+\frac{7}{x^4}}+\sqrt{\frac{4x^2}{x^4}+\frac{3x}{x^4}+\frac{5}{x^4}}}$
= $\lim\limits_{x \to \infty}\frac{-1-\frac{5}{x}+\frac{2}{x^2}}{\sqrt{\frac{3}{x^2}-\frac{2}{x^3}+\frac{7}{x^4}}+\sqrt{\frac{4}{x^2}+\frac{3}{x^3}+\frac{5}{x^4}}}$
= $\frac{-1-0+0}{\sqrt{0-0+0}+\sqrt{0+0+0}}$
= $-\frac{1}{0}$
= $-\infty $
Cara Praktis:
$\lim\limits_{x \to \infty}(\sqrt{3x^2-2x+7}-\sqrt{4x^2+3x+5})$
$a=3$, $p=4$, dan $a < p$ maka:
$\lim\limits_{x \to \infty}(\sqrt{3x^2-2x+7}-\sqrt{4x^2+3x+5})$ = $\infty $

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C. Soal Latihan

1. Tentukan nilai $\lim\limits_{x \to \infty}\frac{(3x-2)^2(5x+7)}{(2x+1)(10x^2+4x-1)}$.
2. Tentukan nilai $\lim\limits_{x \to \infty}\frac{\sqrt{x^2-x+1}}{2x-2}$
3. Tentukan nilai $\lim\limits_{x \to \infty}\frac{3+2^x}{1-2^{x+1}}$
4. $\lim\limits_{x \to \infty}(\sqrt{x(9x+5)}-3x+1)$
5. $\lim\limits_{x \to \infty}(\sqrt{4x+3}-\sqrt{4x-5})\sqrt{x+7}$

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