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:

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

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