# Question 3bf5e

Nov 24, 2016

#### Explanation:

Here is the definition of $| x - 1 |$:

|x - 1| = x - 1; x >= 1
|x - 1| = -x + 1; x < 1#

Because the limit is approaching from the positive side only, we can replace $| x - 1 |$ with (x - 1):

${\lim}_{x \to {1}^{+}} \frac{{x}^{2} - 1}{x - 1}$

Because the expression evaluated at the limit results in the indeterminate form $\frac{0}{0}$, the use of L'Hôpital's rule is warrented.

Take the derivative of the numerator:

$\frac{d \left({x}^{2} - 1\right)}{\mathrm{dx}} = 2 x$

Take the derivative of the denominator:

$\frac{d \left(x - 1\right)}{\mathrm{dx}} = 1$

The rule says that:

${\lim}_{x \to {1}^{+}} \frac{2 x}{1}$

Goes to the same limit as the original expression.

${\lim}_{x \to {1}^{+}} 2 x = 2$

Therefore, the original limit:

${\lim}_{x \to {1}^{+}} \frac{{x}^{2} - 1}{|} x - 1 | = 2$