﻿ Infinite Limits--Examples and interactive practice problems explained step by step when limits do not exist

# Infinite Limits---Skill Practice ### Quick Overview

infinite limits - when limits don't exist because the function becomes infinitely large
1. Infinite limit can only occur when the limit has the form $$\frac n 0$$ for $$n\neq0$$
2. Need to examine the one-sided limits.

### Examples

##### Example 1

Evaluate $$\displaystyle \lim_{x\to 3}\,\frac{x+2}{x-3}$$

Step 1

Determine the form of the limit.

$$\displaystyle\lim_{x\to 3}\,\frac{x+2}{x-3} = \frac{3 + 2}{3-3} = \frac 5 0$$

The limit does not exist, but it has the necessary form so that it might be an infinite limit.

Step 2

Examine the left-hand limit.

1. The numerator approaches 5, so it will be positive.
2. Since $$x$$ is approaching 3 from the left, the denominator will be negative.
3. As the denominator shrinks to 0, the function becomes infinitely large.

Result: $$\displaystyle \lim_{x\to3^-}\,\frac{x+2}{x+3} = -\infty$$

Step 3

Examine the right-hand limit.

1. The numerator approaches 5, so it will be positive.
2. Since $$x$$ is approaching 3 from the right, the denominator will be positive.
3. As the denominator shrinks to 0, the function becomes infinitely large.

Result: $$\displaystyle \lim_{x\to3^+}\,\frac{x+2}{x+3} = \infty$$

$$\displaystyle \lim_{x\to 3}\,\frac{x+2}{x-3}$$ does not exist.

##### Example 2

Evaluate: $$\displaystyle \lim_{x\to-2}\,\frac{x^2-5}{x^2+4x+4}$$

Step 1

Determine the form of the limit.

$$\\ \displaystyle\lim_{x\to-2}\,\frac{x^2-5}{x^2+4x+4} % = \frac{(-2)^2-5}{(-2)^2+4(-2)+4} % = \frac{-1} 0 \\$$

The limit doesn't exist, but it has the $$\frac n 0$$ form so it might be an infinite limit.

Step 2

Try factoring the denominator so the one-sided limits are easier to analyze.

$$\displaystyle\lim_{x\to-2}\,\frac{x^2-5}{x^2+4x+4} % = \displaystyle\lim_{x\to-2}\,\frac{x^2-5}{(x+2)^2}$$

Step 3

Examine the one-sided limits.

1. In both cases, the numerator approaches -1, so the numerator will be negative.
2. In both cases, the denominator is being squared, so it will always be positive.
3. In both cases, the denominator is approaching 0, so the function will become infinitely large.

Both one-sided limits grow infinitely large in the negative direction.

$$\displaystyle \lim_{x\to-2}\, \frac{x^2-5}{x^2+4x+4} = -\infty$$

##### Example 3

Evaluate: $$\displaystyle \lim_{x\to 4}\,\frac{x^2-16}{x^2-8x+16}$$

Step 1

Determine the form of the limit.

$$\displaystyle\lim_{x\to 4} \frac{x^2-16}{x^2-8x+16} % = \frac{(4)^2-16}{x^2-8(4)+16} % = \frac 0 0$$

Step 2

Find and divide out any common factors.

\\ \begin{align*} \displaystyle\lim_{x\to 4}\,\frac{x^2-16}{x^2-8x+16} % & = \displaystyle\lim_{x\to 4}\,\frac{(x-4)(x+4)}{(x-4)^2}\\[6pt] % & = \displaystyle\lim_{x\to 4}\,\frac{x+4}{x-4} \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\displaystyle\lim_{x\to 4}\,\frac{x+4}{x-4} = \frac 8 0$$

We know the limit doesn't exist. Since it has the $$\frac n 0$$ form, it might be an infinite limit.

Step 4

Examine the left-hand limit.

1. The numerator approaches 8, so it will be positive.
2. Since $$x$$ is approaching 4 from the left, the denominator will be negative.
3. As the denominator shrinks to 0, the function will become infinitely large.

So, $$\displaystyle \lim_{x\to 4^-}\,\frac{x+4}{x-4} = -\infty$$

Step 5

Examine the right-hand limit.

1. The numerator approaches 8, so it is positive.
2. Since $$x$$ is approaching 4 from the right, the denominator will be positive.
3. As the denominator shrinks to 0, the function will become infinitely large.

So, $$\displaystyle \lim_{x\to 4^+}\,\frac{x+4}{x-4} = \infty$$

$$\displaystyle \lim_{x\to 4}\, \frac{x^2-16}{x^2-8x+16}$$ does not exist.

##### Example 4

Evaluate: $$\displaystyle \lim_{x\to 5}\,\frac{3x^2+4x}{x^2 - 25}$$

Step 1

Evaluate the limit to determine its form.

$$\displaystyle\lim_{x\to 5}\,\frac{3x^2+4x}{x^2 - 25} % = \frac{3(5)^2+4(5)}{(5)^2 - 25} % = \frac{3(25) + 20}{25-25} % = \frac{95} 0$$

The limit doesn't exist, but it has the $$\frac n 0$$ form so it might be an infinite limit.

Step 2

Examine the left-hand limit.

1. The numerator approaches 95, so it will be positive.
2. Since $$x$$ is approaching 5 from the left, we know $$x<5$$. This means $$x^2 < 25$$. So the denominator will be negative.
3. The denominator is headed to zero, so the function will become infinitely large.

Taken all together, these statements mean

$$\\ \displaystyle \lim_{x\to5^-}\,\frac{3x^2+4x}{x^2-5} = -\infty \\$$

Step 3

Examine the right-hand limit.

1. The numerator approaches 95, so it will be positive.
2. Since $$x$$ is approaching 5 from the right, we know $$x>5$$. This means $$x^2> 25$$. So the denominator will be positive.
3. The denominator is headed to zero, so the function will become infinitely large.

Taken all together, these statements mean

$$\displaystyle \lim_{x\to 5^+}\,\frac{3x^2+4x}{x^2-5} = \infty$$

$$\\ \displaystyle \lim_{x\to 5}\, \frac{3x^2+4x}{x^2 - 25} \\$$ does not exist.

### Practice Problems

Step 1

Determine the form of the limit.

$$\displaystyle \lim_{x\to-3}\,\frac{x-5}{x+3} % = \frac{-3-5}{-3+3} % = \frac{-8} 0$$

The limit doesn't exist, but it might be an infinite limit.

Step 2

Examine the left-hand limit.

1. The numerator approaches -8, so it will be negative.
2. Since $$x$$ is approaching -3 from the left, the denominator will be negative.
3. As the denominator approaches zero, the function will become infinitely large.

Taken all together, the statements tell us

$$\displaystyle \lim_{x\to-3^-}\,\frac{x-5}{x+3} = \infty$$

Step 3

Examine the right-hand limit.

1. The numerator approaches -8, so it will be negative.
2. Since $$x$$ is approaching -3 from the right, the denominator will be positive.
3. As the denominator approaches zero, the function will become infinitely large.

Taken all together, these statements mean

$$\displaystyle \lim_{x\to-3^+}\,\frac{x-5}{x+3} = -\infty$$

$$\displaystyle \lim_{x\to-3}\, \frac{x-5}{x+3}$$ does not exist.

Step 1

Determine the form of the limit.

$$\displaystyle \lim_{x\to1}\,\frac{x^2+4x+3}{x-1} % = \frac{(1)^2+4(1)+3}{1-1} % = \frac 8 0$$

The limit doesn't exist, but it has the right form so it might be an infinite limit.

Step 2

Examine the left-hand limit.

1. The numerator approaches 8, so it will be positive.
2. Since $$x$$ is approaching 1 from the left, the denominator will be negative.
3. As the denominator approaches zero, the function will become infinitely large.

Taken all together, these statements mean

$$\displaystyle \lim_{x\to1^-}\,\frac{x^2+4x+3}{x-1} = -\infty$$

Step 3

Examine the right-hand limit .

1. The numerator approaches 8, so it will be positive.
2. Since $$x$$ is approaching 1 from the right, the denominator will be positive.
3. As the denominator approaches zero, the function will become infinitely large.

Taken all together, these statements mean

$$\displaystyle \lim_{x\to1^+}\,\frac{x^2+4x+3}{x-1} = \infty$$

$$\displaystyle \lim_{x\to1^+}\, \frac{x^2+4x+3}{x-1}$$ does not exist.

Step 1

Determine the form of the limit.

$$\displaystyle\lim_{x\to7}\,\frac{x^2-4}{(x-7)^2} % = \frac{(7)^2-4}{(7-7)^2} % = \frac{45} 0$$

The limit does not exist. However, it has the correct form so it might be an infinite limit.

Step 2

Examine the one-sided limits.

In both cases,

1. The numerator approaches 45, so it will be positive.
2. The denominator is being squared, so it will always be positive.
3. As the denominator approaches zero, the function will become infinitely large.

Take all together, these statements mean both one-sided limits are infinite in the positive direction.

$$\displaystyle \lim_{x\to7}\,\frac{x^2-4}{(x-7)^2} = \infty$$

Step 1

Determine the form of the limit.

$$\displaystyle\lim_{x\to\frac 1 2}\,\frac{6x-9}{4x^2-4x+1} % = \frac{6\left(\frac 1 2\right)-9}{4\left(\frac 1 2\right)^2-4\left(\frac 1 2\right)+1} % = \frac{-6}{4(1/4) - 2 + 1} % = \frac{-6} 0$$

The limit does not exist. However, it has the correct form so it might be an infinite limit.

Step 2

Factor the denominator so it is easier to analyze.

$$\displaystyle\lim_{x\to\frac 1 2}\,\frac{6x-9}{4x^2-4x+1} % = \displaystyle\lim_{x\to\frac 1 2}\,\frac{6x-9}{(2x-1)^2}$$

Step 3

Examine the one-sided limits.

1. The numerator approaches -6, so it will be negative.
2. The denominator is a perfect square, so as $$x$$ approaches $$\frac 1 2$$, the denominator will always be positive.
3. As the denominator approaches 0, the function will become infinitely large.

Taken all together, these statements tell us that both one-sided limits are infinite in the negative direction.

$$\displaystyle \lim_{x\to\frac 1 2}\, \frac{6x-9}{4x^2-4x+1} = -\infty$$.

Step 1

Determine the form of the limit.

$$\displaystyle\lim_{x\to 8}\,\frac{x^2-7x-8}{(x-8)^2} % = \frac{(8)^2 - 7(8) - 8}{(8-8)^2} % = \frac{64 - 56 - 8}{0} % = \frac 0 0$$

Step 2

Find and divide out any common factors.

$$\displaystyle\lim_{x\to 8}\,\frac{x^2-7x-8}{(x-8)^2} % = \displaystyle\lim_{x\to 8}\,\frac{(x-8)(x+1)}{(x-8)^2} % = \displaystyle\lim_{x\to 8}\,\frac{x+1}{x-8}$$

Step 3

Evaluate the simpler limit.

$$\displaystyle\lim_{x\to 8}\,\frac{x+1}{x-8} % = \frac{8+1}{8-8} % = \frac 9 0$$

The limits doesn't exist, but it does have the correct form so it might be an infinite limit.

Step 4

Examine the limit from the left.

1. The numerator approaches 9, so it is positive.
2. Since $$x$$ is approaching 8 from the left, the denominator is negative.
3. As the denominator approaches zero, the function becomes infinitely large.

Taken all together, these statements mean

$$\displaystyle \lim_{x\to8^-}\,\frac{x+1}{x-8} = -\infty$$

Step 5

Examine the right from the right.

1. The numerator approaches 9, so it will be positive.
2. Since $$x$$ is approaching 8 from the right, the denominator is positive.
3. As the denominator approaches zero, the function becomes infinitely large.

Taken together, these statements mean

$$\displaystyle \lim_{x\to8^+}\,\frac{x+1}{x-8} = \infty$$

$$\displaystyle \lim_{x\to 8}\, \frac{x^2-7x-8}{(x-8)^2}$$ does not exist.

Step 1

Determine the form of the limit.

$$\displaystyle \lim_{x\to -\frac 3 2}\,\frac{8x^2+14x+3}{4x^2 + 12x + 9} % = \frac{8\left(-\frac 3 2\right)^2+14\left(-\frac 3 2\right)+3}{4\left(-\frac 3 2\right)^2 + 12\left(-\frac 3 2\right) + 9} % = \frac{18-21+3}{9 - 18 + 9} % = \frac 0 0$$

Step 2

Find and divide out any common factors.

$$\displaystyle \lim_{x\to -\frac 3 2}\,\frac{8x^2+14x+3}{4x^2 + 12x + 9} % = \displaystyle \lim_{x\to -\frac 3 2}\,\frac{(2x+3)(4x+1) }{(2x+3)^2} % = \displaystyle \lim_{x\to -\frac 3 2}\,\frac{4x+1}{2x+3}$$

Step 3

Evaluate the simpler limit.

$$\displaystyle \lim_{x\to -\frac 3 2}\,\frac{4x+1}{2x+3} % = \frac{4\left(-\frac 3 2\right) + 1}{2\left(-\frac 3 2\right) + 3} % = \frac{-6+1}{-3+3} % = \frac{-5} 0$$.

The limit does not exist, but it has the correct form so that it might be an infinite limit.

Step 4

Examine the limit from the left.

1. The numerator approaches -5, so it will be negative.
2. Since $$x$$ is approaching $$-\frac 3 2$$ from the left, the denominator will be negative.
3. As the denominator approaches zero, the function will become infinitely large.

Taken together, these statements mean

$$\displaystyle \,\lim_{x\to-\frac{3}{2}^-} \frac{4x+1}{2x+3} = \infty$$

Step 5

Examine the limit from the right.

1. The numerator approaches -5, so it will be negative.
2. Since $$x$$ is approaching $$-\frac 3 2$$ from the right, the denominator will be positive.
3. As the denominator approaches zero, the function will become infinitely large.

Taken together, these statements mean

$$\displaystyle \,\lim_{x\to-\frac{3}{2}^+} \frac{4x+1}{2x+3} = -\infty$$

$$\displaystyle \lim_{x\to -\frac 3 2}\,\frac{8x^2+14x+3}{4x^2 + 12x + 9}$$ does not exist

Step 1

Determine the form of the limit.

$$\displaystyle \lim_{x\to 2}\,\frac{2x^2-14}{x^2-4} % = \frac{2(2)^2 - 14}{(2)^2-4} % = \frac{-6}{0}$$

The limit doesn't exist, but it has the correct form so it might be an infinite limit.

Step 2

Examine the left-hand limit.

1. The numerator approaches -6, so it will be negative.
2. Since $$x$$ is approaching 2 from the left, we know $$x<2$$ so $$x^2 < 4$$. This means the denominator will be negative.
3. As the denominator approaches zero, the function will become infinitely large.

Taken together, these statements tell us

$$\\ \displaystyle \lim_{x\to2^-}\, \frac{2x^2-14}{x^2-4} = \infty \\$$

Step 3

Examine the right-hand limit.

1. The numerator approaches -6, so it will be negative.
2. Since $$x$$ is approaching 2 from the right, we know $$x>2$$ so $$x^2 > 4$$. This means the denominator will be positve.
3. As the denominator approaches zero, the function will become infinitely large.

Taken together, these statements tell us

$$\displaystyle \lim_{x\to2^+}\, \frac{2x^2-14}{x^2-4} = -\infty$$

$$\\ \displaystyle \lim_{x\to 2} \,\frac{2x^2-14}{x^2-4} \\$$ does not exist.

Step 1

Determine the form of the limit.

$$\displaystyle \lim_{x\to 6}\,\frac{x^2 - 5x - 36}{x^2-36} % = \frac{(6)^2 - 5(6) - 36}{(6)^2-36} % = \frac{- 30}{0}$$

The limit does not exist, but it does have the correct form so it might be an infinite limit.

Step 2

Examine the left-hand limit.

1. The numerator approaches -30, so it will be negative.
2. Since x is approaching 6 from the left, we know $$x<6$$, so $$x^2 < 36$$, so the denominator will be negative.
3. As the denominator approaches zero, the function will become infinitely large.

These statements mean

$$\displaystyle \lim_{x\to 6^-}\, \frac{x^2 - 5x - 36}{x^2-36} = \infty$$

Step 3

Examine the right-hand limit.

1. The numerator approaches $$-30$$, so it will be negative.
2. As $$x$$ approaches 6 from the right, we know $$x>6$$, so $$x^2 > 36$$, so the denominator will be positive.
3. As the denominator approaches zero, the function will become infinitely large.

These statements indicate

$$\displaystyle \lim_{x\to 6^+}\, \frac{x^2 - 5x - 36}{x^2-36} = -\infty$$

$$\displaystyle \lim_{x\to 6}\, \frac{x^2 - 5x - 36}{x^2-36}$$ does not exist. 