Question #3bf5e

1 Answer
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_(xto1^+) (x^2 -1)/(x - 1)

Because the expression evaluated at the limit results in the indeterminate form 0/0, the use of L'Hôpital's rule is warrented.

Take the derivative of the numerator:

(d(x^2 - 1))/dx = 2x

Take the derivative of the denominator:

(d(x - 1))/dx = 1

The rule says that:

lim_(xto1^+) (2x)/1

Goes to the same limit as the original expression.

lim_(xto1^+) 2x = 2

Therefore, the original limit:

lim_(xto1^+) (x^2 -1)/|x - 1| = 2