# Rational Function Limits

## Value of rational functions, and cases when denominator equals zero.

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Rational Function Limits

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
For the rational function \begin{align*}f(x) = \frac{p(x)} {q(x)}\end{align*} and any real number a,
\begin{align*}\lim_{x \rightarrow a} f(x) = \frac{p(a)} {q(a)} \text{ if } q(a) \neq 0\end{align*}.
However, if \begin{align*} \,\! q(a) = 0 \end{align*} then the function may or may not have any outputs that exist.

### 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 one-sided 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 one-sided 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.

1. \begin{align*}\lim_{x \to 1} \frac{-12x^2 + 12}{4x - 4}\end{align*}
2. \begin{align*}\lim_{x \to 2} \frac{\frac{3}{x+3} - \frac{11}{9}}{2x - 4}\end{align*}
3. \begin{align*}\lim_{x \to \frac{-57}{56}} \frac{\frac{-5x - 3}{2x + 3} - \frac{13}{6}}{-56x - 57}\end{align*}
4. \begin{align*}\lim_{x \to 2} \frac{2x^2 - 5x + 2}{-x + 2}\end{align*}
5. \begin{align*}\lim_{x \to \frac{3}{4}} \frac{4x^2 + 5x - 6}{4x - 3}\end{align*}
6. \begin{align*}\lim_{x \to 4} \frac{\frac{-3}{-2x + 3} - \frac{3}{5}}{6x - 24}\end{align*}
7. \begin{align*}\lim_{x \to \frac{-3}{2}} \frac{\frac{-4x - 3}{-2x+2} - \frac{5}{2}} {-2x -3}\end{align*}
8. \begin{align*}\lim_{x \to 4} \frac{x^2 - 8x + 16}{x - 4}\end{align*}
9. \begin{align*}\lim_{x \to\frac{10}{39} } \frac{\frac{3x + 3}{-3x + 4} - \frac{7}{6}} {39x - 10}\end{align*}
10. \begin{align*}\lim_{x \to -4} \frac{3x^2 + 7x - 20}{-x - 4}\end{align*}
11. \begin{align*}\lim_{x \to -4} \frac{4x^2 + 14x -8}{x + 4}\end{align*}
12. \begin{align*}\lim_{x \to \frac{-18}{13}} \frac{\frac{-4x + 1}{-3x + 5} - \frac{5}{7}} {-13x - 18}\end{align*}
13. \begin{align*}\lim_{x \to -2} \frac{\frac{3}{x + 4} - \frac{3}{2}}{-3x - 6}\end{align*}
14. \begin{align*}\lim_{x \to \frac{-3}{4}} \frac{\frac{-x + 3}{4x + 3} - \frac{11}{2}}{-4x - 3}\end{align*}
15. \begin{align*}\lim_{x \to \frac{1}{4}} \frac{-8x^2 - 2x + 1}{-4x + 1}\end{align*}
16. \begin{align*}\lim_{x \to \frac{1}{4}} \frac{16x^2 - 16x + 3}{-4x + 1}\end{align*}
17. \begin{align*}\lim_{x \to 0} \frac{x^2 + 3x}{x}\end{align*}

To see the Review answers, open this PDF file and look for section 8.5.

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### Vocabulary Language: English

TermDefinition
Conjugates Conjugates are pairs of binomials that are equal aside from inverse operations between them, e.g. $(3 + 2x)$ and $(3 - 2x)$.
Continuous Continuity for a point exists when the left and right sided limits match the function evaluated at that point. For a function to be continuous, the function must be continuous at every single point in an unbroken domain.
discontinuous A function is discontinuous if the function exhibits breaks or holes when graphed.
limit A limit is the value that the output of a function approaches as the input of the function approaches a given value.
Rational Function A rational function is any function that can be written as the ratio of two polynomial functions.
rationalization Rationalization generally means to multiply a rational function by a clever form of one in order to eliminate radical symbols or imaginary numbers in the denominator. Rationalization is also a technique used to evaluate limits in order to avoid having a zero in the denominator when you substitute.
theorem A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven.