Question

How do you find the x values at which `f(x)=x/(x^2-x)` is not continuous, which of the discontinuities are removable?

Answer 1

A rational function is continuous on its domain and only on its domain.

Explanation:

The domain of `f` is all real `x` except solutions to `x^2-x=0`.

Since the solutions are `0` and `1`, the function `f` is not continuous at `0` and at `1`.

A discontinuity of function `f` at `a` is removable if and only if `lim_(xrarra)f(x)` exists.

Checking the discontinuity at `0`:

`lim_(xrarr0) x/(x^2-x) = lim_(xrarr0)1/(x-1) = -1`.

This discontinuity is removable.

Checking the discontinuity at `1`:

`lim_(xrarr0) x/(x^2-x) = lim_(xrarr0)1/(x-1)` does not exist.

This discontinuity is not removable.

Additional facts
`lim_(xrarr1^-)1/(x-1) = -oo` and `lim_(xrarr0^+)1/(x-1) = oo`

A discontinuity like this is sometimes called an infinite discontinuity. Infinite discontinuities are not removable.