# What are the asymptotes and removable discontinuities, if any, of f(x)= (x^2+4)/(x-3) ?

Dec 25, 2015

No removable discontinuities, and the 2 asymptotes of this function are $x = 3$ and $y = x$.

#### Explanation:

This function is not defined at $x = 3$, but you can still evaluate the limits on the left and on the right of $x = 3$.

${\lim}_{x \to {3}^{-}} f \left(x\right) = - \infty$ because the denominator will be strictly negative, and ${\lim}_{x \to {3}^{+}} f \left(x\right) = + \infty$ because the denomiator will be strictly positive, making $x = 3$ an asymptote of $f$.

For the 2nd one, you need to evaluate $f$ near the infinities. There is a property of rational functions telling you that only the biggest powers matter at the infinities, so it means that $f$ will be equivalent to ${x}^{2} / x = x$ at the infinites, making $y = x$ another asymptote of $f$.

You can't remove this discontinuity, the 2 limits at $x = 3$ are different.

Here is a graph :
graph{(x^2 + 4)/(x - 3) [-163.5, 174.4, -72.7, 96.2]}