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.
In what follows, \( C \) represents the constant of integration where applicable.
Questions with Detailed Solutions
Click on each question to view its detailed step-by-step solution.
Question 1
Find the limit: \( \displaystyle \lim_{x \to -1} \dfrac{x}{|x|} \)
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The steps to find the limit are:
\[ \lim_{x \to -1} \dfrac{x}{|x|} = \dfrac{-1}{|-1|} = \dfrac{-1}{1} = -1 \]Question 2
Evaluate the limit: \( \displaystyle \lim_{x \to 1} \dfrac{x}{|x|} \)
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The steps to find the limit are:
\[ \lim_{x \to 1} \dfrac{x}{|x|} = \dfrac{1}{|1|} = 1 \]Question 3
Find the limit: \( \displaystyle \lim_{x \to 0} \dfrac{x}{|x|} \)
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Recall that:
\( |x| = x \) for \( x \ge 0 \) and \( |x| = -x \) for \( x < 0 \)
Let us calculate the limit from the left of \( x = 0 \) where \( x < 0 \) and therefore \( |x| = -x \):
\[ \lim_{x \to 0^-} \dfrac{x}{|x|} = \lim_{x \to 0^-} \dfrac{x}{-x} = \lim_{x \to 0^-} (-1) = -1 \]Let us calculate the limit from the right of \( x = 0 \) where \( x > 0 \) and therefore \( |x| = x \):
\[ \lim_{x \to 0^+} \dfrac{x}{|x|} = \lim_{x \to 0^+} \dfrac{x}{x} = \lim_{x \to 0^+} 1 = 1 \]The limits from the left and from the right of \( x = 0 \) are not equal, therefore:
\[ \lim_{x \to 0} \dfrac{x}{|x|} \quad \text{does not exist} \]The graph of \( f(x) = \dfrac{x}{|x|} \) is shown below, illustrating that the limits from the left and right of 0 are not equal.
Question 4
Evaluate the limit: \( \displaystyle \lim_{x \to \infty} \dfrac{x}{|x|} \)
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As \( x \) increases indefinitely, \( x > 0 \) and therefore \( |x| = x \). Hence:
\[ \lim_{x \to \infty} \dfrac{x}{|x|} = \lim_{x \to \infty} \dfrac{x}{x} = \lim_{x \to \infty} 1 = 1 \]Question 5
Find the limit: \( \displaystyle \lim_{x \to -\infty} \dfrac{x}{|x|} \)
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As \( x \) decreases indefinitely, \( x < 0 \) and therefore \( |x| = -x \). Hence:
\[ \lim_{x \to -\infty} \dfrac{x}{|x|} = \lim_{x \to -\infty} \dfrac{x}{-x} = \lim_{x \to -\infty} (-1) = -1 \]Question 6
Does the limit \( \displaystyle \lim_{x \to -2} \dfrac{|x + 2|}{x + 2} \) exist?
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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 \( x + 2 \ge 0 \) or \( x \ge -2 \), then \( |x + 2| = x + 2 \).
If \( x + 2 \le 0 \) or \( x \le -2 \), then \( |x + 2| = -(x + 2) \).
Let us calculate the limit from the left of \( x = -2 \):
\[ \lim_{x \to -2^-} \dfrac{|x + 2|}{x + 2} = \lim_{x \to -2^-} \dfrac{-(x + 2)}{x + 2} = \lim_{x \to -2^-} (-1) = -1 \]Let us calculate the limit from the right of \( x = -2 \):
\[ \lim_{x \to -2^+} \dfrac{|x + 2|}{x + 2} = \lim_{x \to -2^+} \dfrac{x + 2}{x + 2} = \lim_{x \to -2^+} 1 = 1 \]The limits from the left and from the right of \( x = -2 \) are not equal, therefore:
\[ \lim_{x \to -2} \dfrac{|x + 2|}{x + 2} \quad \text{does not exist} \]Question 7
Find the limit: \( \displaystyle \lim_{x \to 1} \dfrac{x^2 + 2x - 3}{|x - 1|} \)
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At \( x = 1 \), both numerator and denominator are equal to zero, so they share a common factor of \( x - 1 \). Factoring the numerator:
\[ \lim_{x \to 1} \dfrac{x^2 + 2x - 3}{|x - 1|} = \lim_{x \to 1} \dfrac{(x - 1)(x + 3)}{|x - 1|} \]Recall that \( |x - 1| = x - 1 \) for \( x \ge 1 \) and \( |x - 1| = -(x - 1) \) for \( x \le 1 \).
Limit from the left of \( x = 1 \):
\[ \lim_{x \to 1^-} \dfrac{x^2 + 2x - 3}{|x - 1|} = \lim_{x \to 1^-} \dfrac{(x - 1)(x + 3)}{-(x - 1)} = \lim_{x \to 1^-} -(x + 3) = -4 \]Limit from the right of \( x = 1 \):
\[ \lim_{x \to 1^+} \dfrac{x^2 + 2x - 3}{|x - 1|} = \lim_{x \to 1^+} \dfrac{(x - 1)(x + 3)}{x - 1} = \lim_{x \to 1^+} (x + 3) = 4 \]The limits from the left and from the right are not equal, therefore:
\[ \lim_{x \to 1} \dfrac{x^2 + 2x - 3}{|x - 1|} \quad \text{does not exist} \]The graph of \( f(x) = \dfrac{x^2 + 2x - 3}{|x - 1|} \) is shown below, clearly showing that the left and right limits at 1 are not equal.
Question 8
Evaluate the limit: \( \displaystyle \lim_{x \to \infty} \dfrac{x^2 + 5x + 7}{|x + 2|} \)
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As \( x \) increases indefinitely, \( x + 2 \) also increases indefinitely and therefore \( x + 2 \ge 0 \), hence:
\[ \lim_{x \to \infty} \dfrac{x^2 + 5x + 7}{|x + 2|} = \lim_{x \to \infty} \dfrac{x^2 + 5x + 7}{x + 2} = +\infty \]Question 9
Find the limit: \( \displaystyle \lim_{x \to -\infty} \dfrac{x^2 + 5x + 7}{|x + 2|} \)
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As \( x \) decreases indefinitely, \( x + 2 \) also decreases indefinitely and therefore \( x + 2 \le 0 \), hence:
\[ \lim_{x \to -\infty} \dfrac{x^2 + 5x + 7}{|x + 2|} = \lim_{x \to -\infty} \dfrac{x^2 + 5x + 7}{-(x + 2)} = +\infty \]The graph of \( f(x) = \dfrac{x^2 + 5x + 7}{|x + 2|} \) is shown below, illustrating that \( y = f(x) \) increases indefinitely as \( x \) increases or decreases indefinitely.
Exercises
Calculate the following limits. Click each exercise to check your answers.
Exercise 1
\( \displaystyle \lim_{x \to 0} \dfrac{x^2}{|x|} \)
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Exercise 2
\( \displaystyle \lim_{x \to -6^-} \dfrac{-(x + 6)}{|x + 6|} \)
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Exercise 3
\( \displaystyle \lim_{x \to -6^+} \dfrac{-(x + 6)}{|x + 6|} \)
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Exercise 4
\( \displaystyle \lim_{x \to 3} \dfrac{x^2 - x - 6}{|x - 3|} \)