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Answer 1 · 2016-09-25T21:16:57

#2#

#lim_(x->5) ((x+5)/(x-5)-100/(x^2-25))#

#= lim_(x->5) ((x+5)/(x-5)-100/((x+5)(x-5)))#

Combining the fractions:
#= lim_(x->5) ((x+5)(x+5)-100)/((x+5)(x-5))#

Plug in #x = 5# and you'll see that this is in indeterminate #0/0# form so we can use L'Hôpital's Rule.

But first we consolidate the numerator and denominator

#= lim_(x->5) (x^2+10 x -75)/(x^2 - 25)#

and by L'Hôpital's Rule

#= lim_(x->5) (2x+10)/(2x)#

As the numerator and denominator are continuous, so is the quotient. So can now plug the value x = 5 directly in:

# = (2(5) + 10)/(2(5)) = 2#

L'Hôpital's Rule, to me, is kinda brute-force meets black-box, so if you need a more refined solution, do ask.