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

Oct 26, 2015

Since ${x}^{2} - 1 = \left(x + 1\right) \left(x - 1\right)$ we will have discontinuities
at $x = + 1 \mathmr{and} x = - 1$ (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 $x = + 1 \mathmr{and} x = - 1$ are vertical asymptotes.

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