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

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