# Evaluate the limit of the fraction (x^2-4)/(x^2+x-6) as x approaches to 2.

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We have to find: lim x-->2 [(x^2-4)/(x^2+x-6)]

substituting x= 2 gives the indeterminate form 0/0. So we can substitute the numerator and the denominator with their derivatives using l'Hopital's rule.

=> lim x-->2 [ 2x / (2x + 1)]

Substitute x = 2

=> 4/ 5

**The required value of lim x-->2 [(x^2-4)/(x^2+x-6)] = 4/5**

First, we'll substitute x by 2 and we'll verify if it is an indetermination:

lim (x^2-4)/(x^2+x-6) = (2^2-4)/(2^2+2-6) = (4-4)/(4+2-6) = 0/0

Since we've get an indetermination, "0/0" type, we'll apply L'Hospital rule:

lim (x^2-4)/(x^2+x-6) = lim (x^2-4)'/(x^2+x-6)'

lim (x^2-4)'/(x^2+x-6)' = lim 2x/(2x+1)

We'll substitute again x by 2:

lim 2x/(2x+1) = 2*2/(2*2+1) = 4/5

**The limit of the fraction, as x approaches to 2, is: lim (x^2-4)/(x^2+x-6) = 4/5.**