# Why doesn't work to evaluate the limit (x^2-1)/(x-1), x-->1, by substituting x=1?

*print*Print*list*Cite

We can not find the limit by substituting 1 because 1 is the root for upper and lower functions. That means that (x-1) is a common factor. By factorizing, we could eliminate the common factor and then simplify the function then we could find the limit.

In this case:

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

Now we substitute 1 :

==> lim (x^2-1)/(x-1) when x-->1 = lim (x+1) = 2

In the expression (x^2-1)/(x-1), as x--> 1, both numerator and denominator approach zero . This makes the expression to be a 0/0 form which is indeterminate. The reason both numerator and denominator have x-1 a, common factor.

To get the limit as x--> 1, we can divide by the common factoe and then take the limit. Another technique is to use L'Hospital's method of diffrentiaing both numerator and denominator and then try for the limit as x-->1.

So dividing both numearator and denominator by x-1, we get:

(x^2-1)/(x-1) = x+1.

So Lt x-->1 of (x^2-1)/(x-1) = Lt x--> 1 { (x+1) } = 1+1 =2.

We cannot simply calculate the lim, by substituting x=1, because the function is not defined in x=1.

We cannot apply the Quotient Law, also, because the limit of denominator is 0, too.

We'll factorize the numerator, being a difference of squares:

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

We'll simplify and we'll get:

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