Evaluate the limit of the function (x^2-1)/(x-1) if x goes to 1?

### 3 Answers | Add Yours

We'll substitute x by 1 in the expression of the function:

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

x->1

Since we've get an indetermination, we can apply L'Hospital's rule:

lim f(x)/g(x) = lim f'(x)/g'(x)

Let f(x) = x^2 - 1 => f'(x) = 2x

Let g(x) = x - 1 => g'(x) = 1

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

x->1 x->1

We'll replace x by 1:

lim 2x/1 = 2/1 = 2

**The requested limit of the given function, if x approaches to 1, is lim (x^2 - 1)/(x-1) = 2.**

Since `(x^2-1) = (x-1)(x+1)`

`lim_(x->1) (x^2-1)/(x-1) = lim_(x->1) ((x+1)(x-1))/(x-1) = lim_(x->1) (x+1) = 2`

The limit `lim_(x->1) (x^2-1)/(x-1)` has to be determined.

If we substitute x = 1 in the polynomial `(x^2-1)/(x-1)` we get `0/0` which is indeterminate.

One way around this is by factorizing the denominator and numerator and canceling common terms.

`lim_(x->1) (x^2-1)/(x-1)`

Use the property `x^2 - y^2 = (x - y)(x + y)`

`lim_(x->1) ((x - 1)(x + 1))/(x-1)`

= `lim_(x->1) (x + 1)`

Now substituting x = 1 gives 1 + 1 = 2

The limit `lim_(x->1) ((x - 1)(x + 1))/(x-1) = 2`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes