Question

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

Accepted Answer

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.

Answer

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

Calculus

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

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