Question

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

Answer 1

`f` has a removable discontinuity at `0`.

Explanation:

Recall that `absx = {(x, "if",x >= 0),(-x, "if",x < 0) :}`

So `f(x) = {(x^2/x, "if",x > 0),(x^2/-x, "if",x < 0) :}`

Notice that `f` is not defined for `x=0`, so `f` is not continuous at `0`.

We can simplify the definition of `f`, to get

`f(x) = {(x, "if",x > 0),(-x, "if",x < 0) :}`

This is simply the absolute value function except that it is left undefined at `0`.

We know (I think) that the absolute value function is continuous, so `f` is continuous everywhere except `0`.

Although `f(0)` is not defined, we can see that `lim_(xrarr0)f(x)` exists. (It equals `0`), so the discontinuity at `0` is removable.