# Limits to Determine Vertical Asymptotes

Recall that the function $f(x) = \frac{1}{x}$ has a vertical asymptote at $x = 0$. We will now precisely define what a vertical asymptote is using limits.

Definition: A function $f$ has a Vertical Asymptote (often abbreviated as V.A.) at $x = a$ if at least one of the following limits are satisfied:a) $\lim_{x \to a} f(x) = \pm \infty$.b) $\lim_{x \to a^-} f(x) = \pm \infty$.c) $\lim_{x \to a^+} f(x) = \pm \infty$. |

We should note from the definition above that only one need to hold for us to know that a vertical asymptote exists at $x = a$. For example, if we know that $\lim_{x \to a^+} f(x) = \infty$ then we know a vertical asymptote exists at $x = a$, and it would be redundant to show whether $\lim_{x \to a^-} f(x) = \pm \infty$. We will now apply this concept into verifying a vertical asymptote of a function.

## Example 1

**Find any vertical asymptotes to the function $f(x) = \frac{1}{x + 2}$.**

We note that as $x \to -2^-$, $f(x) \to -\infty$. Therefore, there exists a vertical asymptote at $x = -2$. Alternative, we could look at the limit from the otherside. As $x \to -2^+$, $f(x) \to \infty$ which implies once again that there exists a vertical asymptote at $x = -2$.

## Example 2

**Find any vertical asymptotes to the function $f(x) = \ln (x)$.**

We note that $f$ is only defined for $x > 0$, so we suspect there exists an asymptote at $x = 0$. We clearly cannot take the limit from the lefthand side, so we will prove there is a vertical asymptote there by evaluating the righthand limit, that is, $\lim_{x \to 0^+} f(x)$. As $x \to 0^+$, $f(x) \to -\infty$ as depicted in the graph below:

Therefore, there exists a vertical asymptote at $x = 0$.