Question

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

Accepted Answer

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

Answer

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

Answer

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

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

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