How do you find the limit of #(x^2+x-12)/(x-2)# as #x->2#?

1 Answer
Alan P.
Oct 12, 2016

#lim_(xrarr2) (x^2+x-12)/(x-2) = color(green)(+-oo)#

Explanation:

At #x=2#
#color(white)("XXX")#the numerator becomes #2^2+2-12= -6#
and
#color(white)("XXX")#the denominator becomes #2-2=0#

L'Hospital's (or L'Hopital's) Rule does not apply. L'Hopital's Rule only applies if both the numerator and denominator are #0#.

The limit of a non-negative numerator by a denominator that is approaching #0# is #+-oo# (depending upon which side the denominator is approaching "0" from).

Here is a graph of the #(x^2+x-12)/(x-2)#.
Note that as #xrarr2# the limit does not approach 5 (as implied by those attempting to use L'Hopital's Rule).

enter image source here

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