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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#