Question

What are all the removable discontinuities/holes and vertical asymptotres for `(2x^2)/(x^2-1)`?

Answer 1

Since `x^2-1=(x+1)(x-1)` we will have discontinuities
at `x=+1andx=-1` (or the denominator will be `=0`

Explanation:

These discontinuities are non-removable, as it makes a difference whether you approach these values from below of from above (try this and see the graph). This means that `x=+1andx=-1` are vertical asymptotes.

Also, when `x` grows bigger (positive or negative) the function starts to look more like
`(2cancel(x^2))/(cancel(x^2))=2` so `y=2` is a horizontal asymptote.
Between the values of `x=+1andx=-1` the function will be negative with a maximum at `(0,0)`, so in the values of the function there is a gap between `0and+2`
graph{2x^2/(x^2-1) [-16.02, 16.01, -8.01, 8.01]}