Limit di Ketakhinggaan Suatu Fungsi

Limit suatu fungsi untuk \(x\to\infty\) adalah nilai limit fungsi dengan nilai x mendekati tak hingga.

Perhatikan pasangan nilai 𝑥 dan \(\dfrac{1}{𝑥}\) berikut, menjelaskan bahwa \(\lim\limits_{x\to\infty}\dfrac{1}{x}=0\)

\(x\) 1 2 5 10 100 1000 1000000 … \(\to\infty\)
\(\dfrac{1}{x}\) 1 0,5 0,2 0,1 0,01 0,001 0,000001 … \(\to{0}\)

Limit di Ketakhinggaan Fungsi Aljabar

Tujuan Pembelajaran:
Menentukan nilai limit di ketakhinggaan fungsi alajabr dalam bentuk fungsi rasional dan irasional (bentuk akar).

Limit di Ketakhinggaan Fungsi Rasional

\(\lim\limits_{x\to{\color{blue}+\infty}}\dfrac{1}{x}=0\) dan \(\lim\limits_{x\to{\color{red}-\infty}}\dfrac{1}{x}=0\)

Contoh 1

Tentukan nilai \(\lim\limits_{x\to{\color{blue}+\infty}}\dfrac{2^{50}}{x}\) dan \(\lim\limits_{x\to{\color{red}-\infty}}\dfrac{2^{2024}}{x}\)

\(\begin{flalign*} \lim\limits_{x\to{\color{blue}+\infty}}\dfrac{2^{50}}{x} &= 2^{50}\times\lim\limits_{x\to{\color{blue}+\infty}}\dfrac{1}{x} \\ &= 2^{50}\times{0}=0 \end{flalign*}\)

\(\begin{flalign*} \lim\limits_{x\to{\color{red}-\infty}}\dfrac{2^{2024}}{x} &= 2^{2024}\times\lim\limits_{x\to{\color{red}-\infty}}\dfrac{1}{x} \\ &= 2^{2024}\times{0}=0 \end{flalign*}\)

\(\lim\limits_{x\to\infty} \dfrac{ax^m +b}{px^n +q} = {\left\{\begin{array}{cll} a\;\cdot\; \infty & \text{; untuk } m \text{ > } n & \\ \dfrac{a}{p} & \text{; untuk } m = n & \\ 0 & \text{; untuk } m \text{ < } n & \end{array}\color{white}\right\}}\)

Contoh 2

Tentukan nilai \(\,\lim\limits_{x\to\infty} \dfrac{6x^5-4}{2x^3+100}\,\), \(\,\lim\limits_{x\to\infty} \dfrac{6x^5-4}{3x^5+81}\,\) dan \(\,\lim\limits_{x\to\infty} \dfrac{2x^3+100}{3x^5+81}\)

\(\begin{flalign*} &\lim\limits_{x\to\infty} \dfrac{6x^5-4}{2x^3+100}\times\dfrac{\frac{1}{x^3}}{\frac{1}{x^3}} \\ &= \lim\limits_{x\to\infty} \dfrac{6x^2-\frac{4}{x^3}}{2+\frac{100}{x^3}} \\ &= \dfrac{\infty-0}{2+0}=\infty \end{flalign*}\)

\(\begin{flalign*} &\lim\limits_{x\to\infty} \dfrac{6x^5-4}{3x^5+81}\times\dfrac{\frac{1}{x^5}}{\frac{1}{x^5}} \\ &= \lim\limits_{x\to\infty} \dfrac{6-\frac{4}{x^5}}{3+\frac{81}{x^5}} \\ &= \dfrac{6-0}{3+0}=\dfrac{6}{3}=2 \end{flalign*}\)

\(\begin{flalign*} &\lim\limits_{x\to\infty} \dfrac{2x^3+100}{3x^5+81}\times\dfrac{\frac{1}{x^5}}{\frac{1}{x^5}} \\ &= \lim\limits_{x\to\infty} \dfrac{\frac{2}{x^2}+\frac{100}{x^5}}{3+\frac{81}{x^5}} \\ &= \lim\limits_{x\to\infty} \dfrac{0+0}{3+0} =0 \end{flalign*}\)

wave

Limit di Ketakhinggaan Fungsi Bentuk akar

\(\lim\limits_{x\to\infty} \left( \sqrt{ax+b}-\sqrt{px+q} \right)\)

\(= \lim\limits_{x\to\infty} \left( \sqrt{ax+b}-\sqrt{px+q} \right) \times \dfrac{ \sqrt{ax+b} + \sqrt{px+q}}{ \sqrt{ax+b} + \sqrt{px+q}}\)

\(= \lim\limits_{x\to\infty} \dfrac{ ax+b-(px+q)}{ \sqrt{ax+b} + \sqrt{px+q}}\)

\(= \lim\limits_{x\to\infty} \dfrac{ (a-p)x+(b-q)}{ \sqrt{ax+b} + \sqrt{px+q}} = {\left\{ \begin{array}{ll} 0 & \text{; untuk } a = p \\ (a-p)\cdot\infty & \text{; untuk } a \neq p \end{array} \color{white}\right\}}\)

Contoh 3

Tentukan nilai \(\lim\limits_{x\to\infty} \left( \sqrt{4x-16}-\sqrt{4x+9}\right)\)

\(\small\begin{flalign*}&\lim\limits_{x\to\infty} \left( \sqrt{4x-16}-\sqrt{4x+9} \right) \\ &= \lim\limits_{x\to\infty} \left( \sqrt{4x-16}-\sqrt{4x+9} \right) \times \dfrac{ \sqrt{4x-16} + \sqrt{4x+9}}{ \sqrt{4x-16} + \sqrt{4x+9}} \\ &= \lim\limits_{x\to\infty} \dfrac{ 4x-16-4x-9}{ \sqrt{4x-16} + \sqrt{4x+9}} \\ & = \lim\limits_{x\to\infty} \dfrac{-25}{ \sqrt{4x-16} + \sqrt{4x+9}} = 0\end{flalign*}\)

Cara lain

Diketahui \(a=4\), \(b=-16\), \(p=4\), dan \(q=9\).

Karena \(a=4=p\), maka

\(\small\lim\limits_{x\to\infty} \left( \sqrt{4x-16}-\sqrt{4x+9} \right) = 0\)

\(\lim\limits_{x\to\infty} \left( \sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r} \right)\)

\(= \lim\limits_{x\to\infty} \left( \sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r} \right) \times \dfrac{ \sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}}{ \sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}}\)

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

\(= \lim\limits_{x\to\infty} \dfrac{ (a-p)x^2+(b-q)x+(c-r)}{ \sqrt{ax^2+bx+c} + \sqrt{px^2+qx+r}} = {\left\{ \begin{array}{ll} \dfrac{b-q}{2\sqrt{a}} & \text{; untuk } a = p \\ (a-p)\cdot\infty & \text{; untuk } a \neq p \end{array} \color{white}\right\}}\)

Contoh 4

Tentukan nilai \(\lim\limits_{x\to\infty} \left( \sqrt{x^2+x}-\sqrt{x^2-5x} \right)\)

\(\small\begin{flalign*} &\lim\limits_{x\to\infty} \left( \sqrt{x^2+x}-\sqrt{x^2-5x} \right) \\ &= \lim\limits_{x\to\infty} \left( \sqrt{x^2+x+1}-\sqrt{x^2-5x} \right) \times \dfrac{\sqrt{x^2+x} + \sqrt{x^2-5x}}{\sqrt{x^2+x} + \sqrt{x^2-5x}} \\ &= \lim\limits_{x\to\infty} \dfrac{x^2+x - \left( x^2-5x \right)}{\sqrt{x^2+x}+\sqrt{x^2-5x}} \\ &= \lim\limits_{x\to\infty} \dfrac{1-(-5)}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{5}{x}}}\times\frac{x}{x} \\ &= \dfrac{1-(-5)}{\sqrt{1+0} + \sqrt{1+0}}= \dfrac{1-(-5)}{2\sqrt{1}}=\dfrac{6}{2}=3 \end{flalign*}\)

Cara lain

Diketahui \(a=1\), \(b=1\), \(c=0\), \(p=1\), \(q=-5\), dan \(r=0\).

Karena \(a=1=p\), maka

\(\small\begin{flalign*} &\lim\limits_{x\to\infty} \left( \sqrt{x^2+x}-\sqrt{x^2-5x} \right) \\ &= \frac{b-q}{2\sqrt{a}} = \frac{1-(-5)}{2\sqrt{1}} = \frac{6}{2}=3 \end{flalign*}\)

Tentukan nilai \(\lim\limits_{x\to\infty} \left( \sqrt{9x^2+6x+1}-3x+ 2 \right)\)

\(\begin{flalign*} &\lim\limits_{x\to\infty} \left( \sqrt{9x^2+6x+1}-3x+ 2 \right) \\ &= 2 + \lim\limits_{x\to\infty} \left( \sqrt{9x^2+6x+1}-3x \right) \\ &= 2 + \lim\limits_{x\to\infty} \left( \sqrt{9x^2+6x+1}-\sqrt{9x^2} \right) \\ &= 2 + \dfrac{6-0}{2\sqrt{9}}=2+\dfrac{6}{6}=2+1=3 \end{flalign*}\)

Tentukan nilai \(\lim\limits_{x\to\infty} \left( 2x - 3 - \sqrt{4x^2-16x} \right)\)

\(\begin{flalign*} &\lim\limits_{x\to\infty} \left( 2x - 3 - \sqrt{4x^2-16x} \right) \\ &= -3 + \lim\limits_{x\to\infty} \left( 2x-\sqrt{4x^2-16x} \right) \\ &= -3 + \lim\limits_{x\to\infty} \left( \sqrt{4x^2}-\sqrt{4x^2-16x} \right) \\ &= -3 + \dfrac{0-(-16)}{2\sqrt{4}} = -3 + \dfrac{16}{4}=-3+4=1 \end{flalign*}\)

Tentukan nilai \(\lim\limits_{x\to\infty} \left( 3x - \sqrt{4x^2-8x} - \sqrt{x^2-6x}\right)\)

\(\begin{flalign*} &\lim\limits_{x\to\infty} \left( 3x - \sqrt{4x^2-8x} - \sqrt{x^2-6x}\right) \\ &\lim\limits_{x\to\infty} \left( 2x - \sqrt{4x^2-8x} + x - \sqrt{x^2-6x}\right) \\ &= \lim\limits_{x\to\infty} \left( \sqrt{4x^2} - \sqrt{4x^2-8x} + \sqrt{x^2} - \sqrt{x^2-6x}\right) \\ &= \dfrac{0-(-8)}{2\sqrt{4}} + \dfrac{0-(-6)}{2\sqrt{1}} \\ &= \dfrac{8}{4} + \dfrac{6}{2}=2+3 = 7 \end{flalign*}\)

wave

Limit di Ketakhinggaan Fungsi Trigonometri

  1. \(\lim\limits_{x\to{\infty}} \sin{\frac{1}{x}} = \sin{0} = 0\)
  2. \(\lim\limits_{x\to{\infty}} \cos{\frac{1}{x}} = \cos{0} = 1\)
  3. \(\lim\limits_{x\to{\infty}} x\sin{\frac{1}{x}} = \lim\limits_{x\to{\infty}} \dfrac{1}{\frac{1}{x}}\sin{\frac{1}{x}}\)
    \(= \lim\limits_{x\to{\infty}} \dfrac{\sin{\frac{1}{x}}}{\frac{1}{x}} \qquad{\small\color{olive}\text{Misalkan }p = \frac{1}{x}\text{, sehingga } x\to{\infty} \equiv p\to{0}}\)
    \(= \lim\limits_{p\to{0}} \dfrac{\sin{p}}{p} = 1\)
  4. \(\lim\limits_{x\to{\infty}} \cos^{-1}{\frac{1}{x}} = \cos^{-1}{0} = \frac{\pi}{2}\)

Contoh 5

Tentukan nilai \(\lim\limits_{x\to\infty} 2x \sin{\frac{1}{4x}}\)

Misalkan \(p=\frac{1}{x}\), sehingga \(x\to{\infty} \equiv p\to{0}\)

\(\begin{flalign*} \lim\limits_{x\to\infty} 2x\,\sin{\frac{1}{4x}} &= \lim\limits_{x\to\infty} \frac{2}{\frac{1}{x}}\,\sin{\left(\frac{1}{4}\cdot\frac{1}{x}\right)} \\ &= 2\times\lim\limits_{x\to\infty} \frac{\sin{\left(\frac{1}{4}\cdot\frac{1}{x}\right)}}{\frac{1}{x}} \\ &= 2\times\lim\limits_{p\to{0}} \frac{\sin{\frac{1}{4}p}}{p} \\ &= 2\times\frac{1}{4}=\frac{1}{2} \end{flalign*}\)

wave

Soal Jawab Limit Fungsi di Ketakhinggaan

\(\text{01a. } \lim\limits_{x\to\infty} \dfrac{4x^2(2x^4-x^3+x-1)}{(2x^3+3x-5)^2} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to\infty} \frac{4x^2(2x^4-x^3+x-1)}{(2x^3+3x-5)^2} &= \lim\limits_{x\to\infty} \frac{4x^2(2x^4)}{(2x^3)^2} \\ &= \lim\limits_{x\to\infty} \frac{8x^6}{4x^6} \\ &= \frac{8}{4} = 2 \end{flalign*}\)

\(\text{01b. } \lim\limits_{x\to\infty} \dfrac{6x^3-3x^2+3x-1}{\sqrt{4x^6-6x^4+10}} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to\infty} \dfrac{6x^3-3x^2+3x-1}{\sqrt{4x^6-6x^4+10}} &= \lim\limits_{x\to\infty} \dfrac{6x^3}{\sqrt{4x^6}} \\ &= \lim\limits_{x\to\infty} \dfrac{6x^3}{2x^3} \\ &= \frac{6}{2} = 3 \end{flalign*}\)

\(\text{02a. } \lim\limits_{x\to\infty} \left( \sqrt{16x^2 + 15x -1} - \sqrt{16x^2-9x+1} \right)= \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to\infty} \left( \sqrt{16x^2 + 15x -1} - \sqrt{16x^2-9x+1} \right) &= \frac{15-(-9)}{2\sqrt{16}} & {\color{brown}\small a=16=p} \\ &=\frac{24}{2\times{4}} \\ &= 3 \end{flalign*}\)

\(\text{02b. } \lim\limits_{x\to\infty} \left( 5x+1-\sqrt{25x^2 - 10x + 1} \right)= \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to\infty} \left( 5x+1-\sqrt{25x^2 - 10x +1} \right) &= 1 + \lim\limits_{x\to\infty} \left( \sqrt{25x^2}-\sqrt{25x^2 - 10x +1} \right) \\ &= 1+\frac{0-(-10)}{2\sqrt{25}} \\ &=1+\frac{10}{2\times{5}} \\ &= 1+1=2 \end{flalign*}\)

\(\text{02c. } \lim\limits_{x\to\infty} \left(\sqrt{49x^2 + 14x + 1} - 7x - 5 \right)= \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to\infty} \left( \sqrt{49x^2 + 14x +1} -7x-5 \right) &= -5 + \lim\limits_{x\to\infty} \left( \sqrt{49x^2 + 14x +1} - \sqrt{49x^2} \right) \\ &= -5+\frac{14-0}{2\sqrt{49}} \\ &=-5+\frac{10}{2\times{7}} \\ &= -5+1=-4 \end{flalign*}\)

\(\text{03a. } \lim\limits_{x\to\infty} 2x^2\,\sin{\frac{1}{4x^2}} = \cdots.\)

Misalkan \(p=\frac{1}{x^2}\), sehingga \(x\to{\infty} \equiv p\to{0}\)

\(\begin{flalign*} \lim\limits_{x\to\infty} 2x^2 \sin{\frac{1}{4x^2}} &= \lim\limits_{x\to\infty} \frac{2}{\frac{1}{x^2}} \sin{\left(\frac{1}{4}\cdot\frac{1}{x^2}\right)} \\ &= 2\times\lim\limits_{x\to\infty} \frac{\sin{\left(\frac{1}{4}\cdot\frac{1}{x^2}\right)}}{\frac{1}{x^2}} \\ &= 2\times\lim\limits_{p\to{0}} \frac{\sin{\frac{1}{4}p}}{p} \\ &= 2\times{\frac{1}{4}}=\frac{1}{2} \end{flalign*}\)

\(\text{03b. } \lim\limits_{x\to\infty} 2x^2\,\sin^2{\frac{1}{4x}} = \cdots.\)

Misalkan \(p=\frac{1}{x}\), sehingga \(x\to{\infty} \equiv p\to{0}\)

\(\begin{flalign*} \lim\limits_{x\to\infty} 2x^2 \sin^2{\frac{1}{4x}} &= \lim\limits_{x\to\infty} \frac{2}{\left(\frac{1}{x}\right)^2} \sin^2{\left(\frac{1}{4}\cdot\frac{1}{x}\right)} \\ &= 2\times\lim\limits_{x\to\infty} \frac{\sin^2{\left(\frac{1}{4}\cdot\frac{1}{x}\right)}}{\left(\frac{1}{x}\right)^2} \\ &= 2\times\lim\limits_{p\to{0}} \frac{\sin^2{\frac{1}{4}p}}{p^2} \\ &= 2\times\lim\limits_{p\to{0}} \frac{\sin{\frac{1}{4}p}}{p}\times\lim\limits_{p\to{0}} \frac{\sin{\frac{1}{4}p}}{p} \\ &= 2\times{\frac{1}{4}}\times{\frac{1}{4}}=\frac{1}{8} \end{flalign*}\)

\(\begin{flalign*} \text{04. } &\text{Apakah fungsi } y=f(x) \text{ kontinu pada } x = 2 \text{, } \\ &\text{jika } f(x) = {\left\{\begin{array}{cl} 2x-2 & \text{; untuk } x \text{ < } 2 \\ 2 & \text{; untuk } x = 2 \\ x^2-2 & \text{; untuk } x \text{ > } 2 \end{array}\color{white}\right.} \end{flalign*}\)

Periksa syarat kekontinuan fungsi \(y=f(x)\) di \(x=2\):

  1. \(f(2) = 2\,\) terdefinisi (syarat ke-1 terpenuhi)
  2. \(\lim\limits_{x\to{2}^{-}} f(x) = \lim\limits_{x\to{2}} (2x-2) = 2(2)-2 = 2\)
    dan
    \(\lim\limits_{x\to{2}^{+}} f(x) = \lim\limits_{x\to{2}} (x^2-2) = 2^2-2 = 4-2=2\)
    Diperoleh Limit kiri = limit kanan (syarat ke-2 terpenuhi)
  3. \(\lim\limits_{x\to{2}} f(x) = 2 = f(2)\) (syarat ke-3 terpenuhi)

Karena ketiga syarat kekontinuan fungsi \(\,y=f(x)\,\) di \(\,x=2\,\) terpenuhi, maka \(\,y=f(x)\,\) kontinu di \(\,x=2\). \(\begin{flalign*} \end{flalign*}\)

\(\begin{flalign*}\text{05. } &\text{Tuliskan persamaan asimtot datar } \text{dan asimtot tegak dari fungsi } \\ &f(x) = \dfrac{4x^2-1}{x^2-x-2} \end{flalign*}\)

Asimtot datarnya

\(\begin{flalign*} y &= \lim\limits_{x\to{\infty}} \frac{4x^2-1}{x^2-x-2} \\ &= \frac{4}{1} \\ &= 4 \end{flalign*}\)

Asimtot tegaknya

\(\begin{aligned} x^2-x-2 &= 0 \\ (x+1)(x-2) &= 0 \\ x+1 &= 0 & x-2 &= 0 & \\ x &= -1 & x &= 2 & \end{aligned}\)

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