Question #3bf5e

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Answer 1 · 2016-11-24T18:13:57

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:

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