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 1
Find the limit:
\[
\lim_{x \to -1} \frac{x}{|x|}
\]
Solution
The steps to find the limit are:
\[
\lim_{x \to -1} \frac{x}{|x|}
= \frac{-1}{|-1|}
= \frac{-1}{1}
= -1
\]
Question 2
Evaluate the limit:
\[
\lim_{x \to 1} \frac{x}{|x|}
\]
Solution
The steps to find the limit are
\[
\lim_{x \to 1} \frac{x}{|x|}
= \frac{1}{|1|}
= 1
\]
Question 2
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.
Figure 1. Graph of f(x) = x / |x|.
Question 4
Evaluate the limit:
\[
\lim_{x \to \infty} \frac{x}{|x|}
\]
Solution
as \( x \) increases indefinitely , \( x \gt 0 \) 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 5
Find the limit:
\[
\lim_{x \to -\infty} \frac{x}{|x|}
\]
Solution
as \( x \) decreases indefinitely , \( x \lt 0 \) 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 6
Does the limit
\[
\lim_{x \to - 2} \frac{|x + 2|}{x + 2}
\]
exists?
Let us calculate the limit from the left of \( -2 \) (where \( x \le -2 \)) and from the right of \( -2 \) (where \( x \ge -2 \)) separately.
Recall that:
If \( x + 2 \ge 0 \quad \text{or} \quad x \ge -2 \), then
\[
|x + 2| = x + 2
\]
and
If \( x + 2 \le 0 \quad \text{or})\quad x \le -2 \), 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
\[
\lim_{x \to - 2} \frac{|x + 2|}{x + 2} \; \; \text{does not exist}
\]
\[
\lim_{x \to 1} \frac{x^2+2x-3}{|x - 1|} = \frac{(1)^2+2(1)-3}{|(1) - 1|} = \dfrac{0}{0} \; \; \text{indeterminate}
\]
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
\[ | x - 1 | = x - 1 \quad \text{for} \quad x - 1 \ge 0 \quad \text{or} \quad x \ge 1 \]
and
\[ | x - 1 | = - (x - 1) \quad \text{for} \quad x - 1 \le 0 \quad \text{or} \quad x \le 1 \]
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 \( f(x) =\dfrac{x^2 + 2 x - 3}{|x - 1|} \) is shown below and we clearly see that the limits from the left and right of 1 are not equal.
Figure 2. Graph of \( f(x) =\dfrac{x^2 + 2 x - 3}{|x - 1|} \).
As \( x \) decreases indefinitely, \( x + 2\) also decreases indefinitely and therefore \( x + 2 \le 0 \), hence
\[
\lim_{x \to -\infty} \dfrac{x^2+5x+7}{|x + 2|} = \lim_{x \to -\infty} \frac{x^2+5x+7}{-(x + 2)} = + \infty
\]
The graph of \( f(x) = \dfrac{x^2+5x+7}{|x + 2|} \) is shown below and we clearly see that \( y = f(x) \) increases indefinitely as x increases indefinitely and also as \( x \) decreases indefinitely.
Figure 3. Graph of \( f(x) = \dfrac {x^2 + 5 x + 7}{|x + 2|} \).
