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Question

Find the horizontal and vertical asymptotes of the graph of the function. (You need not sketch the graph.) f(x)=x22x24f(x)=\frac{x^{2}-2}{x^{2}-4}

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Let's solve the exercise:

First, we will try to find vertical asymptotes\textbf{vertical asymptotes} using the rule for Finding Vertical Asymptotes of Rational Functions\textbf{Finding Vertical Asymptotes of Rational Functions}:

P(x)=x22Q(x)=x24Q(x)=0x24=0x2=4x1,2=±2\begin{align*} P(x)&=x^2-2 \\ Q(x)&=x^2-4 \\ \\ Q(x)&=0 \\ x^2-4&=0 \\ x^2&=4 \\ \color{#c34632}x_{1,2}&\color{#c34632}=\pm 2 \\ \end{align*}

We can conclude that x=2x=-2 and x=2x=2 are vertical asymptotes\textbf{vertical asymptotes}. (P(±2)0P(\pm2)\ne0)

Now, let's try to find horizontal asymptotes\textbf{horizontal asymptotes}. We need to find a limit either when xx \to \infty or xx\to -\infty:

limx(x22x24)=limxx2x22x2x2x24x2Divide numerator and denominator by x2.=1010Calculate.=1Simplify.\begin{align*} \lim_{x\to\infty} \left(\frac{x^2-2}{x^2-4}\right) &=\lim_{x\to\infty} \frac{\frac{x^2}{x^2}-\frac{2}{x^2}}{\frac{x^2}{x^2}-\frac{4}{x^2}} && \text{Divide numerator and denominator by $x^2$.}\\ &=\frac{1-0}{1-0} && \text{Calculate.}\\ &\color{#c34632}=1 && \text{Simplify.}\\ \end{align*}

We can conclude that y=1y=1 is a horizontal asymptote\textbf{horizontal asymptote}.

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