What is #lim x-> -1# of #(x^2+2x+1)/(x^2+3x+2)#?

1 Answer
Daniel L.
Jul 9, 2016

#lim_{x->-1}(x^2+2x+1)/(x^2+3x+2)=0#

Explanation:

First we can find that #-1# is the root of both numerator and denominator. So the limit can be written as:

#lim_{x->-1}(x^2+2x+1)/(x^2+3x+2)=lim_{x->-1}((x+1)^2)/((x+1)(x+2))#

This can also be written as:

#lim_{x->-1}(x+1)/(x+2)=0/1=0#

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