How do you evaluate the limit of #(x^2+5x+6)/(x^2+x-2)# as #x->2#?

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Answer 1 · 2017-12-29T12:30:40

See below.

#lim_ (x->2)((x+2)(x+3))/((x+2)(x-1))#

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

#=5/1=5#

Answer 2 · 2017-12-30T01:16:10

#lim_(x to 2)(x^2+5x+6)/(x^2+x-2) =5#

Direct substitution works in this case:

#lim_(x to 2)(x^2+5x+6)/(x^2+x-2) = (4+10+6)/(4+2-2) = 20/4=5#