Finding the Asymptotes of a Rational Function Using Limits
Find any asymptotes of the rational function \(R(x)=\dfrac{3x^{2}-12}{ 2x^{2}-9x+10}\).
The domain of \(R\) is \(\left\{ x |\ x\neq \dfrac{5}{2}\hbox{ and }x\neq 2\right\} \). Since \(R\) is a rational function, it is continuous on its domain, that is, all real numbers except \(x=\dfrac{5}{2}\) and \(x=2\).
To check for vertical asymptotes, we find the limits as \(x\) approaches \( \dfrac{5}{2}\) and \(2\). First we consider \(\lim\limits_{x\rightarrow \frac{ 5}{2}^{-}}R(x)\). \begin{equation*} \lim\limits_{x\rightarrow \frac{5}{2}^{-}}R(x)=\!\!\lim\limits_{x\rightarrow \frac{5}{2}^{-}}\left[ \frac{3(x-2)(x+2)}{(2x-5)(x-2)}\right] =\!\!\lim\limits_{x\rightarrow \frac{5}{2}^{-}}\left[ \frac{3(x+2)}{(2x-5)} \right] =3\!\!\lim\limits_{x\rightarrow \frac{5}{2}^{-}}\frac{x+2}{2x-5}=-\infty \end{equation*}
That is, as \(x\) approaches \(\dfrac{5}{2}\) from the left, \(R\) becomes unbounded in the negative direction.
The graph of \(R\) has a vertical asymptote at \(x=\dfrac{5}{2}\).
To determine the behavior to the right of \(x=\dfrac{5}{2}\), we find the right-hand limit. \begin{equation*} \lim\limits_{x\rightarrow \frac{5}{2}^{+}}R(x)=\lim\limits_{x\rightarrow \frac{5}{2}^{+}}\left[ \frac{3(x+2)}{(2x-5)}\right] = 3\lim\limits_{x\rightarrow \frac{5}{2}^{+}}\frac{x+2}{2x-5}=\infty \end{equation*}
As \(x\) approaches \(\dfrac{5}{2}\) from the right, the graph of \(R\) becomes unbounded in the positive direction.
Next we consider \(\lim\limits_{x\rightarrow 2}R(x)\). \begin{equation*} \lim\limits_{x\rightarrow 2}R(x)=\lim\limits_{x\rightarrow 2}\frac{ 3(x-2)(x+2)}{(2x-5)(x-2)}=\lim\limits_{x\rightarrow 2}\frac{3(x+2)}{2x-5}= \frac{3(2+2)}{2\cdot 2-5}=\frac{12}{-1}=-12 \end{equation*}
The function \(R\) does not have a vertical asymptote at \(2.\)
Since \(2\) is not in the domain of \(R,\) the graph of \(R\) has a hole at the point \((2,-12)\).
To check for horizontal asymptotes, we find the limits at infinity. \[ \begin{eqnarray*} \lim_{x\rightarrow \infty }R( x) &=&\lim_{x\rightarrow \infty } \frac{3x^{2}-12}{2x^{2}-9x+10}\underset{\underset{\underset{\color{#0066A7}{\hbox{and denominator by \(2x^{2}\)}}}{\color{#0066A7}{\hbox{Divide the numerator}}}}{\color{#0066A7}{\uparrow}}} {=} \lim_{x\rightarrow \infty }\frac{\dfrac{3}{2}-\dfrac{6}{x^{2}}}{1-\dfrac{9}{ 2x}+\dfrac{5}{x^{2}}}=\frac{\lim\limits_{x\rightarrow \infty }\left( \dfrac{3 }{2}-\dfrac{6}{x^{2}}\right) }{\lim\limits_{x\rightarrow \infty }\left( 1- \dfrac{9}{2x}+\dfrac{5}{x^{2}}\right) }\\ &=&\dfrac{\dfrac{3}{2}-0}{1-0+0}=\frac{3 }{2} \\ \end{eqnarray*} \]
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\[ \begin{eqnarray*} \lim_{x\rightarrow \infty }R( x) &=&\lim_{x\rightarrow \infty } \frac{3x^{2}-12}{2x^{2}-9x+10}\underset{\underset{\underset{\color{#0066A7}{\hbox{and denominator by \(2x^{2}\)}}}{\color{#0066A7}{\hbox{Divide the numerator}}}}{\color{#0066A7}{\uparrow}}} {=} \lim_{x\rightarrow \infty }\frac{\dfrac{3}{2}-\dfrac{6}{x^{2}}}{1-\dfrac{9}{ 2x}+\dfrac{5}{x^{2}}}\\ &=&\frac{\lim\limits_{x\rightarrow \infty }\left( \dfrac{3 }{2}-\dfrac{6}{x^{2}}\right) }{\lim\limits_{x\rightarrow \infty }\left( 1- \dfrac{9}{2x}+\dfrac{5}{x^{2}}\right) }=\frac{3 }{2} \end{eqnarray*} \]
The line \(y=\dfrac{3}{2}\) is a horizontal asymptote of the graph of \(R\) for \(x\) unbounded in the negative direction and for \(x\) unbounded in the positive direction.