How do you evaluate the limit #(x^2-x-6)/(x+2)# as x approaches #-2#?

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Answer 1 · 2016-10-29T21:06:16

You may do the division or use L'Hopital's Rule both tell you that it goes to -5.

#lim_(xto-2)(x^2 - x - 6)/(x + 2) = #

Do the division:

#lim_(xto-2)x - 3 = -5#

Or use L'Hopital's rule:

#lim_(xto-2)(2x - 1)/1 = -5#

Answer 2 · 2016-10-29T21:11:42

First of all...

#(x^2-x-6)/(x+2)#

#=((x+2)(x-3))/(x+2)#

#=x-3#

Now, in the limit as x approaches minus 2, x-3 will approach -5.

So your answer is -5 .