Question

What are the discontinuities of `f(x) = (x^4 - 1)/(x-1)`?

Answer 1

The only real discontinuity is when the denominator `x-1=0`
But I think you call that a hole. The value of `y`, either approached from above or below `x=1` give the same limit.

You can rewrite the function:

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

and cancel out the `(x-1)`'s ( only if `x!=1`)

`=(x^2+1)(x+1)` which has no further discontinuities
graph{(x^4-1)/(x-1) [-10, 10, -5, 5]}