Question

What are the removable and non-removable discontinuities, if any, of `f(x)=(x^2 - 3x + 2)/(x^2 - 1)`?

Answer 1

Removable at `x=1` and non-removable at `x=-1`

Explanation:

`f(x)=(x^2 - 3x + 2)/(x^2 - 1)` is a rational function.

Therefore, `f` is continuous on its domain.

The domain of `f` is `RR - {-1,1}`.

So `f` has discontinuities at `-1` and at `1`. (It is continuous everywhere else.)

In order to determine whether a discontinuity is removable, we need to look at the limit of `f` as `x` approaches the discontinuity.

At `x=1`
`lim_(xrarr1)f(x) = lim_(xrarr1)(x^2 - 3x + 2)/(x^2 - 1)`.

This limit has indeterminate form `0/0`, so we factor and simplify.

`lim_(xrarr1)f(x) = lim_(xrarr1)((x-2)(x-1))/((x+1)(x-1)`

`= lim_(xrarr1)(x-2)/(x+1) = -1/2`

Because the limit exists, the discontinuity is removable.

At `x=-1`
`lim_(xrarr-1)f(x) = lim_(xrarr1)(x^2 - 3x + 2)/(x^2 - 1)`.

This limit has form `6/0`, so the limit does not exist and the discontinuity cannot be removed.