Question

What are the asymptotes and removable discontinuities, if any, of `f(x)=(x^2+1)/(x^2-1)`?

Answer 1

asymptotes occur at `x = 1 and x=-1`

Explanation:

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

first factor the denominator, it is the difference of squares:

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

so the removable discontinuities are any factors that cancel out, since the numerator is not factorable there are no terms that cancel out, therefore, the function has no removable discontinuities.

so both factors in the denominator are asymptotes, set the denominator equal to zero and solve for x:

`(x+1)(x-1)=0`

`x = 1 and x=-1`

so the asymptotes occur at `x = 1 and x=-1`

graph{(x^2 + 1)/(x^2-1) [-10, 10, -5, 5]}