# Using l'Hospital theorem evaluate the limit of (x^2+11x-12)/(x-1), for x->1.

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We have to find the value of lim x--> 1[ (x^2+11x-12)/(x-1)].

As substituting x = 0 gives us the indeterminate form 0/0, we can use l'Hopital's theorem and substitute the numerator and the denominator with their derivatives.

We get :

lim x--> 1 [2x + 11]

substitute x = 1

=> 2 + 11

=> 13

**The value of lim x-->1 [(x^2+11x-12)/(x-1)] = 13.**

We know that l'Hospital theorem could be applied if the limit gives an indetermination.

We'll verify if the limit exists, for x = 1.

We'll substitute x by 1 in the expression of the function.

lim y = lim (x^2+11x-12)/(x-1)

lim (x^2+11x-12)/(x-1) = (1+11-12)/(1-1) = 0/0

We've get an indetermination case.

We could solve the problem in 2 ways, at least.

We'll apply L'Hospital rule:

lim f(x)/g(x) = lim f'(x)/g'(x)

f(x) = x^2+11x-12 => f'(x) = 2x+11

g(x) = x-1 => g'(x) = 1

lim (x^2+11x-12)/(x-1) = lim (2x+11)

lim (x^2+11x-12)/(x-1) = 2*1 + 11

**lim (x^2+11x-12)/(x-1) = 13**