Finding the limit of a rational function is actually much less complex than it may seem, in fact many of the limits you have already evaluated have been rational functions.
In this lesson, you will gain more experience working with rational function limits, and will use another theorem which simplifies the process of finding the limit of some rational functions.
Rational Function Limits
Sometimes finding the limit of a rational function at a point a is difficult because evaluating the function at the point a leads to a denominator equal to zero. The box below describes finding the limit of a rational function.
Theorem: The Limit of a Rational Function


Examples
Example 1
Find \begin{align*}\lim_{x \rightarrow 3} \frac{2  x} {x  2}\end{align*}.
Using the theorem above, we simply substitute x = 3: \begin{align*}\lim_{x \rightarrow 3} \frac{2  x} {x  2} = \frac{2  3} {3  2} = 1\end{align*}
Example 2
Find \begin{align*}\lim_{x \rightarrow 3} \frac{x + 1} {x  3}\end{align*}.
Notice that the domain of the function is continuous (defined) at all real numbers except at x = 3. If we check the onesided limits we see that \begin{align*}\lim_{x \rightarrow 3^{+}} \frac{x + 1} {x  3}=\infty\end{align*} and \begin{align*}\lim_{x \rightarrow 3^{}} \frac{x + 1} {x  3}=\infty\end{align*}. Because the onesided limits are not equal, the limit does not exist.
Example 3
Find \begin{align*}\lim_{x \rightarrow 2} \frac{x^2  4} {x  2}\end{align*}.
Notice that the function here is discontinuous at x = 2, that is, the denominator is zero at x = 2. However, it is possible to remove this discontinuity by canceling the factor x  2 from both the numerator and the denominator and then taking the limit:
\begin{align*}\lim_{x \rightarrow 2} \frac{x^2 4} {x  2} = \lim_{x \rightarrow 2} \frac{(x  2)(x + 2)} {x  2} = \lim_{x \rightarrow 2} (x + 2) = 4\end{align*} 

This is a common technique used to find the limits of rational functions that are discontinuous at some points. When finding the limit of a rational function, always check to see if the function can be simplified.
Example 4
Find \begin{align*}\lim_{x \rightarrow 3} \frac{2x  6} {x^2 + x  12}\end{align*}.
The numerator and the denominator are both equal to zero at x = 3, but there is a common factor x  3 that can be removed (that is, we can simplify the rational function):
\begin{align*}\lim_{x \rightarrow 3} \frac{2x  6} {x^2 + x  12}\end{align*}  \begin{align*}= \lim_{x \rightarrow 3} \frac{2(x  3)} {(x + 4) (x  3)}\end{align*}  

\begin{align*}= \lim_{x \rightarrow 3} \frac{2} {x + 4}\end{align*}  
\begin{align*}= \frac{2} {7}\end{align*} 
Example 5
Find \begin{align*}\lim_{x \to 1} \frac{5x^2 +x +4}{x  1}\end{align*}.
\begin{align*}\frac{(5x  4)(x  1)}{(x  1)}\end{align*} ..... Start by factoring the numerator
Since we have (x  1) in both numerator and denominator, we know that the original function is equal to just \begin{align*}5x 4\end{align*} except where it is undefined (1).
Therefore the closer we get to inputing 1, the closer we get to the same value, whether from the + or  side.
To find the value, just solve \begin{align*}5x  4\end{align*} for \begin{align*}x = 1\end{align*}.
\begin{align*}\therefore \lim_{x \to 1} \frac{5x^2 +x +4}{x  1} = 5 \cdot 1  4 \to 9\end{align*}
Example 6
Find \begin{align*}\lim_{x \to 2} \frac{x^2 + 2x + 8}{x + 2}\end{align*}.
\begin{align*}\frac{(x + 4)(x + 2)}{(x + 2)}\end{align*} ..... Start by factoring the numerator
Since we have (x + 2) in both numerator and denominator, we know that the original function is equal to just \begin{align*}x + 4\end{align*} except where it is undefined (2).
Therefore the closer we get to substituting 2, the closer we get to the same output value, whether from the + or  side.
To find the value, just solve \begin{align*}x + 4\end{align*} for \begin{align*}x = 2\end{align*}.
\begin{align*}\therefore \lim_{x \to 2} \frac{x^2 + 2x + 8}{x + 2} = (2) + 4 \to 6\end{align*}
Review
Solve the following rational function limits.
 \begin{align*}\lim_{x \to 1} \frac{12x^2 + 12}{4x  4}\end{align*}
 \begin{align*}\lim_{x \to 2} \frac{\frac{3}{x+3}  \frac{11}{9}}{2x  4}\end{align*}
 \begin{align*}\lim_{x \to \frac{57}{56}} \frac{\frac{5x  3}{2x + 3}  \frac{13}{6}}{56x  57}\end{align*}
 \begin{align*}\lim_{x \to 2} \frac{2x^2  5x + 2}{x + 2}\end{align*}
 \begin{align*}\lim_{x \to \frac{3}{4}} \frac{4x^2 + 5x  6}{4x  3}\end{align*}
 \begin{align*}\lim_{x \to 4} \frac{\frac{3}{2x + 3}  \frac{3}{5}}{6x  24}\end{align*}
 \begin{align*}\lim_{x \to \frac{3}{2}} \frac{\frac{4x  3}{2x+2}  \frac{5}{2}} {2x 3}\end{align*}
 \begin{align*}\lim_{x \to 4} \frac{x^2  8x + 16}{x  4}\end{align*}
 \begin{align*}\lim_{x \to\frac{10}{39} } \frac{\frac{3x + 3}{3x + 4}  \frac{7}{6}} {39x  10}\end{align*}
 \begin{align*}\lim_{x \to 4} \frac{3x^2 + 7x  20}{x  4}\end{align*}
 \begin{align*}\lim_{x \to 4} \frac{4x^2 + 14x 8}{x + 4}\end{align*}
 \begin{align*}\lim_{x \to \frac{18}{13}} \frac{\frac{4x + 1}{3x + 5}  \frac{5}{7}} {13x  18}\end{align*}
 \begin{align*}\lim_{x \to 2} \frac{\frac{3}{x + 4}  \frac{3}{2}}{3x  6}\end{align*}
 \begin{align*}\lim_{x \to \frac{3}{4}} \frac{\frac{x + 3}{4x + 3}  \frac{11}{2}}{4x  3}\end{align*}
 \begin{align*}\lim_{x \to \frac{1}{4}} \frac{8x^2  2x + 1}{4x + 1}\end{align*}
 \begin{align*}\lim_{x \to \frac{1}{4}} \frac{16x^2  16x + 3}{4x + 1}\end{align*}
 \begin{align*}\lim_{x \to 0} \frac{x^2 + 3x}{x}\end{align*}
Review (Answers)
To see the Review answers, open this PDF file and look for section 8.5.