Question

Evaluate the limit of y=(2x^2-x-3)/(x+1), x-->-1 using L'Hopital theorem.

Accepted Answer

We have to find the value of lim x--> -1 [(2x^2-x-3)/(x+1)]
If we substitute x = -1 in (2x^2-x-3)/(x+1) we get 0 / 0. So L'Hopital's theorem can be applied. We now have:
lim x--> -1 [(4x -1)/(1)]
substituing x = -1, we get (-4 - 1)/1
=> -5
Limit = -5

Answer

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 (2x^2-x-3)/(x+1)
lim (2x^2-x-3)/(x+1) =  (2+1-3)/(-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) = 2x^2-x-3 => f'(x) = 4x-1
g(x) = x+1 => g'(x) = 1
lim (2x^2-x-3)/(x+1) = lim (4x-1)
lim (2x^2-x-3)/(x+1) = 4*(-1) - 1 = -5

Calculus

Evaluate the limit of y=(2x^2-x-3)/(x+1), x-->-1 using L'Hopital theorem.

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