Calculate the limit of the function f(x)=(x^2-1)/(x-1) x-->1.

First, we'll check to see if we have an indetermination
case. For this reason, we'll substitute x by 1.

lim
(x^2-1)/(x-1)= (1^2 - 1) / (1-1) = 0/0

"0/0" is an
indetermination, so we could use l'Hospital rule.

If lim
(f/g) = 0/0, then lim (f/g)= lim (f')/(g')

We'll
calculate  (x^2-1)':

 (x^2-1)' = 2x -
0

 (x^2-1)' = 2x

We'll
calculate  (x-1)':

 (x-1)' = 1 -
0

 (x-1)' = 1

lim f(x) = lim
(x^2-1)' / (x-1)'

lim (x^2-1)' / (x-1)' = lim 2x /
1

We'll substitute x by 1:

lim
2x / 1 = 2*1/1 = 2

lim (x^2-1)/ (x-1) =
2

Another method to calculate
the limit:

We'll write the differemce of
square from the numerator as:

(x^2-1)=
(x-1)*(x+1)

We'll substitute the difference of squares by
the product:

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

We'll reduce like
terms:

lim (x-1)*(x+1)/ (x-1) = lim
(x+1)

We'll substitute x by
1:

lim (x+1) =
1+1

lim (x+1) =
2