15.5. Limit of a Rational Function

Limit of a Rational Function
A rational function may have a restricted value at x = c such that finding the limit is not straightforward. The rules are listed as follows:

1) Determine the restricted values for the domain of the function.
To find these values, set the denominator to 0 and find the roots of the resulting equation.

Example
f(x) = 3/(x - 4)

When x = 4, the function is undefined, implying that the graph has an infinite discontinuity such that the limit does not exist as x approaches 4. The line x=4 is a vertical asymptote of the graph.

2) If c is a restricted value, the limit may or may not exist.To determine whether it exists, simplify the rational expression. The limit exists if the simplified form is no longer a rational expression or the denominator is no longer zero when x = c.

3) Plug in c after you simplify the expression. If the denominator is still zero, find the one-sided limits of the function by plugging in values that are close to c from the left and right. If the one-sided limits are equal, the limit equals the left- and right-hand limits. Otherwise the limit does not exist.

4) If c is not in the restricted domain, plug it into the rational expression as you usually would.

Finding the Limit Through Direct Substitution
In each rational function below, the value of c is not in the restricted domain so the limit can be found by plugging in the value of c.

Examples
Find the indicated limit.

$1) \lim_{x\rightarrow -3}\, \frac{2x}{3x-1}$

Calculator solution
Type in: lim [ x = -3 ] ( ( 2x ) / (3x - 1 ) )

$2) \lim_{x\rightarrow 2^+} \frac{3-x}{3x+1}$

Calculator solution
Type in: lim [ x = 2 + ] ( ( 3 - x ) / ( 3x + 1 ) )

$3) \lim_{x\rightarrow 1000000} \frac{1}{(x-1)^2}$

Calculator solution
Type in: lim [ x = 1000000 ] ( 1 / ( x - 1 )^2 )

$4) \lim_{x\rightarrow 2} \frac{1}{(x-1)^2}$

Calculator solution
Type in: lim [ x = 2 ] ( 1 / ( x - 1 )^2 )

Limit at a Restricted Value of X
In each rational function below, the value of c is a restricted value of the function's domain. There are three cases that could happen:

Case 1: The Limit Exists
The limit exists at the restricted value if the original rational function can be simplified to cancel out the denominator. At x = c, the graph has a hole.

Illustrative Example
Consider the function f(x) = (x^3 - 4x^2 + x + 6) / (x - 2) and find the limit as x approaches 2. The function is undefined at x = 2, but the expression can be simplified to f(x) = x^2 - 2x - 3. Because the denominator cancels out, the limit as x approaches 2 exists.

Note that finding the limit of f(x) = (x^3 - 4x^2 + x + 6) / (x - 2) gives the same result as finding the limit of f(x) = x^2 - 2x - 3 as shown below.

The graph below shows that the graph still has a hole at x = 2.

More Examples

Evaluate the limit of each rational function as x approaches c where c is a restricted value of the domain.

$1) \lim_{x\rightarrow 2}\frac{x^2+4x-12}{x^2-2x}$

Calculator solution
Type in: lim [ x = 2 ] ( ( x^2 + 4x - 12 ) / ( x^2 - 2x ) )

$2) \lim_{t\rightarrow 5}\frac{t^3-6t^2+25}{t-5}$

Calculator solution
Type in: lim [ t = 5 ] ( ( t^3 - 6t^2 + 25 ) / ( t - 5 ) )

$3) \lim_{x\rightarrow 1}\frac{2-2x^2}{x-1}$

Calculator solution
Type in: lim [ x = 1 ] ( ( 2 - 2x^2 ) / ( x - 1 ) )

$4) \lim_{x\rightarrow -6}\frac{\frac{2x+8}{x^2-12}-\frac{1}{x}}{x+6}$

Calculator solution
Type in: lim [ x = -6 ] ( ( ( 2x + 8 ) / ( x^2 - 12 ) - 1 / x ) / ( x + 6 ) )

$5) \lim_{h\rightarrow 0}\frac{2(-3+h)^2-18}{h}$

Calculator solution
Type in: lim [ h = 0 ] ( ( 2 ( -3 + h )^2 - 18 ) / h )

$6) \lim_{x\rightarrow 0}\frac{\frac{1}{3+x}-\frac{1}{3-x}}{x}$

Calculator solution
Type in: lim [ x = 0 ] ( ( ( 1 / ( 3 + x ) - 1 / ( 3 - x ) ) ) / x )

Case 2: Limit to Infinity
The limit may approach infinity if the denominator is still zero at x = c. As a result, the graph of the function continues increasing or decreasing as x approaches c from the left and right. At this value of x, the graph has an infinite discontinuity. The limit exists if the left- and right-hand limits increase or decrease in the same direction.

Illustrative Example
Consider the function f(x) = 1/x^2 and find the limit as x approaches 0. Because the denominator is zero when we plug in x = 0, we have to evaluate the function from the left and right. The left- and right-hand limits both approach positive infinity, so the limit of f(x) as x approaches 0 is ∞.

More Examples

$1) \lim_{x\rightarrow 0}\frac{6}{x^2}$

Calculator solution
Type in: lim [ x = 0 ] ( 6 / x^2 )

$2) \lim_{x\rightarrow 1}\frac{1}{(x-1)^2}$

Calculator solution
Type in: lim [ x = 1 ] ( 1 / ( x - 1 )^2 )

$3) \lim_{x\rightarrow 1}\frac{x-2}{(x-1)^2}$

Calculator solution
Type in: lim [ x = 1 ] ( ( x - 2 ) / ( x - 1 )^2 )

$4) \lim_{x\rightarrow 0}\frac{6-3x}{2x^2}$

Calculator solution
Type in: lim [ x = 0 ] ( ( 6 - 3x ) / ( 2x^2 ) )

$5) \lim_{x\rightarrow -4}\frac{x}{(2x+8)^2}$

Calculator solution
Type in: lim [ x = - 4 ] ( x / ( 2x + 8 )^2 )

Case 3: The Limit Does Not Exist (DNE)
When the denominator is zero at x = c, x = c is a vertical asymptote of the graph. The function either continues increasing or decreasing as x approaches c from the left and right. If the left- and right-hand limits are in different directions, the limit does not exist.

Illustrative Example
Consider the limit of f(x) = 1/x as x approaches 0. The function is undefined at x = 0, as shown below. Because the left-hand limit is negative infinity while the right-hand limit is positive infinity, the limit as x approaches 0 does not exist.