# Limits of Absolute Value Functions Questions

How to find the limits of absolute value functions; several examples and detailed solutions are presented along with graphical interpretations. A set of exercises with answers is presented at the bottom of the page.

## Find the limits of the following functions.

### Question

Find the limit:

Solution

The steps to find the limit are:

### Question

Evaluate the limit:

Solution

The steps to find the limit are

### Question

Find the limit:
\(
\lim_{x \to 0} \frac{x}{|x|}
\)

Solution

\(
\lim_{x \to 0} \frac{x}{|x|} = \frac{0}{|0|} \; \; \text{indeterminate form}
\)

Recall that

\( | x | = x \) for \( x \le 0 \) and \( | x | = - x \) for \( x \lt 0 \)

Let us calculate the limit from the left of \( x = 0 \) where \( x \lt 0 \) and therefore \( | x | = - x \)

\(
\lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{- x} = \lim_{x \to 0^-} - 1 = -1
\)

Let us calculate the limit from the right of \( x = 0 \) where \( x \gt 0 \) and therefore \( | x | = x \)

\(
\lim_{x \to 0^+} \frac{x}{|x|} = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} 1 =1
\)

The limits from the left and from the right of x = 0 are not equal, therefore

\(
\lim_{x \to 0} \frac{x}{|x|} \; \; \text{does not exist}
\)

The graph of f(x) = x / |x| is shown below and we clearly see that the limits from the left and right of 0 are not equal.

### Question

Evaluate the limit:
\(
\lim_{x \to \infty} \frac{x}{|x|}
\)

Solution

as x increases indefinitely , x > 0 and therefore |x| = x

Hence

\(
\lim_{x \to \infty} \frac{x}{|x|} = \lim_{x \to \infty} \frac{x}{x} = \lim_{x \to \infty} 1 = 1
\)

### Question

Find the limit:
\(
\lim_{x \to -\infty} \frac{x}{|x|}
\)

Solution

as x decreases indefinitely , x < 0 and therefore |x| = - x

Hence

\(
\lim_{x \to -\infty} \frac{x}{|x|} = \lim_{x \to -\infty} \frac{x}{ - x} = \lim_{x \to \infty} - 1 = - 1
\)

### Question

Does the limit
\(
\lim_{x \to - 2} \frac{|x + 2|}{x + 2}
\)
exists?

Solution

\(
\lim_{x \to - 2} \frac{|x + 2|}{x + 2} = \lim_{x \to - 2} \frac{|- 2 + 2|}{- 2 + 2} = \dfrac{0}{0} \;\; \text{indeterminate}
\)

Let us calculate the limit from the left of -2 where x ≤ - 2 and from the right of -2 where x ≥ - 2 separateley.

Recall that

If x + 2 ≥ 0 or x ≥ - 2 then | x + 2 | = x + 2

and

If x + 2 ≤ 0 or x ≤ - 2 then | x + 2 | = - \( ( x + 2 ) \)

Let us calculate the limit from the left of x = - 2.

\(
\lim_{x \to - 2^-} \frac{|x + 2|}{x + 2} = \lim_{x \to -2^-} \frac{-(x + 2)}{x + 2} = \lim_{x \to -2^-} - 1 = -1
\)

Let us calculate the limit from the right of x = - 2

\(
\lim_{x \to - 2^+} \frac{|x + 2|}{x + 2} = \lim_{x \to - 2^+} \frac{x + 2}{x + 2} = \lim_{x \to -2^+} 1 = 1
\)

The limits from the left and from the right of x = - 2 are not equal, therefore

\(
\lim_{x \to - 2} \frac{|x + 2|}{x + 2} \; \; \text{does not exist}
\)

### Question

Find the limit:
\(
\lim_{x \to 1} \frac{x^2+2x-3}{|x - 1|}
\)

Solution

\(
\lim_{x \to 1} \frac{x^2+2x-3}{|x - 1|} = \frac{(1)^2+2(1)-3}{|(1) - 1|} = \dfrac{0}{0} \; \; \text{indeterminate}
\)

At x = 1 both numerator and denominator are equal to zero, they therefore have a common factor x - 1. We factor the numerator.

\(
\lim_{x \to 1} \frac{x^2+2x-3}{|x - 1|} = \lim_{x \to 1} \frac{(x-1)(x+3)}{|x - 1|}
\)

Recall that

| x - 1 | = x - 1 for x - 1 ≥ 0 or x ≥ 1

and

| x - 1 | = - \( ( x - 1 ) \) for x - 1 < 0 or x < 1

Let us calculate the limit from the left of x = 1

\(
\lim_{x \to 1^-} \frac{x^2+2x-3}{|x - 1|} = \lim_{x \to 1^-} \frac{(x-1)(x+3)}{-(x-1)} = \lim_{x \to 1^-} - (x + 3) = - 4
\)

Let us calculate the limit from the right of x = 1

\(
\lim_{x \to 1^+} \frac{x^2+2x-3}{|x - 1|} = \lim_{x \to 1^+} \frac{(x-1)(x+3)}{x-1} = \lim_{x \to 1^-} (x + 3) = 4
\)

The limits from the left and from the right are not equal, therefore

\(
\lim_{x \to 1} \frac{x^2+2x-3}{|x - 1|} \; \; \text{does not exist}
\)

The graph of f(x) = (x^2 + 2 x - 3)/|x - 1| is shown below and we clearly see that the limits from the left and right of 1 are not equal.

### Question

Evaluate the limit:
\(
\lim_{x \to \infty} \frac{x^2+5x+7}{|x + 2|}
\)

Solution

As x increases indefinitely, x + 2 also increases indefinitely and therefore x + 2 > 0, hence

\(
\lim_{x \to \infty} \frac{x^2+5x+7}{|x + 2|} = \lim_{x \to \infty} \frac{x^2+5x+7}{x + 2} = + \infty
\)

### Question

Find the limit:
\(
\lim_{x \to - \infty} \frac{x^2+5x+7}{|x + 2|}
\)

Solution

As x decreases indefinitely, x + 2 also decreases indefinitely and therefore x + 2 < 0, hence

\(
\lim_{x \to -\infty} \frac{x^2+5x+7}{|x + 2|} = \lim_{x \to -\infty} \frac{x^2+5x+7}{-(x + 2)} = + \infty
\)

The graph of \( f(x) = (x^2 + 5 x + 7)/|x + 2| \) is shown below and we clearly see that \( y = f(x) \) increases indefinitely as x increases indefinitely and also as x decreases indefinitely.

## Exercises

Find the limits

1) \(
\lim_{x \to 0} \frac{x^2}{|x|}
\)

2) \(
\lim_{x \to - 6^-} \frac{-(x + 6)}{|x + 6|}
\)

3) \(
\lim_{x \to - 6^+} \frac{-(x + 6)}{|x + 6|}
\)

4) \(
\lim_{x \to 3} \frac{x^2-x-6}{|x - 3|}
\)

## Answers to Above Exercises

1) 0

2) 1

3) - 1

4) does not exist

### More References and links

Calculus Tutorials and Problems - LimitsFind Limits of Functions in Calculus

Introduction to Limits in Calculus

Properties of Limits of Mathematical Functions in Calculus

limits of basic functions

Questions and Answers on Limits in Calculus