Find the limits of the following functions.

Question

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

Solution

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

Question

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

Solution

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

Question

\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.





Question

\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

\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

\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}

Question

\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.





Question

\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

\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.

Exercises

Answers to Above Exercises

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