Question

What are the removable and non-removable discontinuities, if any, of `f(x)=(x^4+4x^2+2x)/(x^2-2x)`?

Answer 1

There is a removable discontinuity at `0` and a non-removable discontinuity at `2`.

Explanation:

`f(x)= (x^4+4x^2+2x)/(x^2-2x)` is a rational function.

Therefore, `f` is continuous on its domain.

The domain of `f` is `RR - {0,2}`.

So `f` has discontinuities at `0` and at `2`. (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=0`
`lim_(xrarr0)f(x) = lim_(xrarr0)(x^4+4x^2+2x)/(x^2-2x)`.

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

`lim_(xrarr0)f(x) = lim_(xrarr0)(x(x^3+4x+2))/(x(x-2))`

`= lim_(xrarr0)(x^3+4x+2)/(x-2) = -1`

Because the limit exists, the discontinuity is removable.

At `x=2`
`lim_(xrarr2)f(x) = lim_(xrarr2)(x^3 +4x^2 + 2x)/(x^2 - 2x)`.

This limit has form `28/0`, so the limit does not exist and the discontinuity cannot be removed.