<meta http-equiv="refresh" content="1; url=/nojavascript/"> Rational Function Limits ( Read ) | Analysis | CK-12 Foundation
Dismiss
Skip Navigation

Rational Function Limits

%
Progress
Practice
Progress
%
Practice Now
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.

Watch This

James Sousa: Ex 3: Determine Limits Analytically by Factoring

Guidance

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

Example A

Find \lim_{x \rightarrow 3} \frac{2 - x} {x - 2} .

Solution:

Using the theorem above, we simply substitute x = 3: \lim_{x \rightarrow 3} \frac{2 - x} {x - 2} = \frac{2 - 3} {3 - 2} = -1

Example B

Find \lim_{x \rightarrow 3} \frac{x + 1} {x - 3} .

Solution:

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 \lim_{x \rightarrow 3^{+}} \frac{x + 1} {x - 3}=\infty and \lim_{x \rightarrow 3^{-}} \frac{x + 1} {x - 3}=-\infty . Because the one-sided limits are not equal, the limit does not exist.

Example C

Find \lim_{x \rightarrow 2} \frac{x^2 - 4} {x - 2} .

Solution:

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:

\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

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.

Vocabulary

A rational function is any function which can be written as the ratio of two polynomial functions.

A discontinuous function exhibits breaks or holes when graphed.

Guided Practice

1) Find \lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12} .

2) Find \lim_{x \to 1} \frac{-5x^2 +x +4}{x - 1} .

3) Find \lim_{x \to -2} \frac{-x^2 + 2x + 8}{x + 2} .

Answers

1) 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):

\lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12} = \lim_{x \rightarrow 3} \frac{2(x - 3)} {(x + 4) (x - 3)}
= \lim_{x \rightarrow 3} \frac{2} {x + 4}
= \frac{2} {7}

2) To find \lim_{x \to 1} \frac{-5x^2 +x +4}{x - 1}

\frac{(-5x - 4)(x - 1)}{(x - 1)} ..... 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 -5x -4 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 -5x - 4 for x = 1 .
\therefore \lim_{x \to 1} \frac{-5x^2 +x +4}{x - 1} = -5 \cdot 1 - 4 \to -9

3) To find \lim_{x \to -2} \frac{-x^2 + 2x + 8}{x + 2}

\frac{(-x + 4)(x + 2)}{(x + 2)} ..... 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 -x + 4 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 -x + 4 for x = -2 .
\therefore \lim_{x \to -2} \frac{-x^2 + 2x + 8}{x + 2} = -(-2) + 4 \to 6

Explore More

Solve the following rational function limits.

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

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Rational Function Limits.

Reviews

Please wait...
Please wait...

Original text