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

By substitution :

lim f(x) = 0/0 (method has failed because 1 is a root for both numerator and denominator

Let us factor:

lim f(x) = lim (x-1)(x+1) /(x-1)

= lim (x+1)

Then,

lim f(x) when x--> 1 = (1+1) = 2

**Then the limit is 2**

To calculate the limit of f(x) = (x^2-1)/(x-1)x-->1.

Solition:

lt f(x) = lt (x^2-1)/(x-1) as x--> 1 isof the form (1^2-1)/(1-1) = 0/0 form of indetermination.

So we can go for L Hospitals rule of differentiating numerator and denominator and then taking limits . We can also go for dividing both numerator and denominator by the common factor and then take limit.

We shall go by the second method.

Numerator = x^2-1 = (x+1)(x-1) and

denominators (x-1).

So x-1 is the common factor.

So lt f(x) = lt(x^2-1)/(x-1) = lt(x+1)(x-1)/(x-1) = lt(x+1) as x --> 1 = (1+1) =2.

So Lt {(x^2-1)/(x-1) ,as x--> 1 } = 2

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**