Question

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

Answer 1

Vertical asymntote at `x=0`, Slant asymtote `y=x`, No removable discontinuties

Explanation:

`f(x) =(x^2-4)/x = x-4/x`

Consider, `lim_(x->0^+) x-4/x =-oo`

And `lim_(x->0^-) x-4/x =+oo`

Hence, `f(x)` has a vertical asymtote at `x=0`

Also, `lim_(x->oo) -4/x = 0`

Hence, `f(x)` has a slant asymtote of `y=x`

We can see both asymtotes on the graph of `f(x)` below.

graph{((x^2-4)/x-y)(x-y)=0 [-18.04, 18, -9, 9.02]}

`f(x)` has no removable discontinuties,