How do you find the limit of  (abs(x+2)-3)/(x-7 )  as x approaches 7?

Oct 9, 2016

${\lim}_{x \to {7}^{+}} \frac{\left\mid x + 2 \right\mid - 3}{x - 7} = \infty$

${\lim}_{x \to {7}^{-}} \frac{\left\mid x + 2 \right\mid - 3}{x - 7} = - \infty$

Explanation:

${\lim}_{x \to 7} \frac{\left\mid x + 2 \right\mid - 3}{x - 7}$

$= \frac{{\lim}_{x \to 7} \left(\left\mid x + 2 \right\mid - 3\right)}{{\lim}_{x \to 7} x - 7}$

and because the numerator is continuous

$= \frac{6}{{\lim}_{x \to 7} x - 7}$

Note that this gives rise to a singularity and a 2 sided limit.

To test this in your head, mentally plug in, say, $x = 6.9$ and $x = 7.1$ So the right-sided limit is positive and the left-sided limit is negative

So we say that:

${\lim}_{x \to {7}^{+}} \frac{\left\mid x + 2 \right\mid - 3}{x - 7} = \infty$

${\lim}_{x \to {7}^{-}} \frac{\left\mid x + 2 \right\mid - 3}{x - 7} = - \infty$

Oct 9, 2016

The limit does not exist.

Explanation:

Simple substitution of $x = 7$ gives a numerator of 6 and a denominator of 0. There would be division of zero.

However, if $x$ approaches 7 from the left, then the expression tends to negative infinity. We write

${\lim}_{x \to {7}^{-}} \frac{\left\mid x + 2 \right\mid - 3}{x - 7} = - \infty$

Similarly, if $x$ approaches 7 from the right, then the expression tends to infinity.

${\lim}_{x \to {7}^{+}} \frac{\left\mid x + 2 \right\mid - 3}{x - 7} = \infty$