Limits of Absolute Value Functions Questions

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

Find the limits of the following functions.

  1. Question
  2. \lim_{x \to - 1} \frac{x}{|x|}

    Solution

    \lim_{x \to - 1} \frac{x}{|x|} = \frac{ - 1}{|- 1|} = \frac{ - 1}{1} = - 1

  3. Question
  4. \lim_{x \to 1} \frac{x}{|x|}

    Solution

    \lim_{x \to 1} \frac{x}{|x|} = \frac{1}{|1|} = 1

  5. Question
  6. \lim_{x \to 0} \frac{x}{|x|}

    Solution

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

    Recall that
    | x | = x for x ≥ 0 and | x | = - x for x < 0
    Let us calculate the limit from the left of x = 0 (x < 0 , | 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 (x > 0 , | 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.
    graph of y = x / |x|

    Figure 1. Graph of f(x) = x / |x|.

  7. Question
  8. \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

  9. Question
  10. \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

  11. Question
  12. \lim_{x \to - 2} \frac{|x + 2|}{x + 2}

    Solution

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

    Recall that
    | x + 2 | = x + 2 for x + 2 ≥ 0 or x ≥ - 2
    and
    | x + 2 | = - ( x + 2) for x + 2 < 0 or x < - 2
    Let us calculate the limit from the left of x = - 2 (x < - 2 , | x + 2 | = - ( 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 (x > - 2 , | x + 2 | = ( 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}

  13. Question
  14. \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 (x < 1 , | x - 1 | = - ( 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 (x > 1 , | x - 1 | = ( 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.
    graph of f(x) = (x^2 + 2 x - 3)/|x - 1|

    Figure 2. Graph of f(x) = (x2 + 2 x - 3)/|x - 1|.

  15. Question
  16. \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

  17. Question
  18. \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) = (x2 + 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.
    graph of f(x) = (x<sup>2</sup> + 2 x - 3)/|x - 1|

    Figure 2. Graph of f(x) = (x2 + 5 x + 7)/|x + 2|.


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 - Limits
Find 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