# How do you find a vertical asymptote for y = sec(x)?

Sep 25, 2014

The vertical asymptotes of $y = \sec x$ are

$x = \frac{\left(2 n + 1\right) \pi}{2}$, where $n$ is any integer,

which look like this (in red).

Let us look at some details.

$y = \sec x = \frac{1}{\cos x}$

In order to have a vertical asymptote, the (one-sided) limit has to go to either $\infty$ or $- \infty$, which happens when the denominator becomes zero there.

So, by solving

$\cos x = 0$

$R i g h t a r r o w x = \pm \frac{\pi}{2} , \pm \frac{3 \pi}{2} , \pm \frac{5 \pi}{2} , \ldots$

$R i g h t a r r o w x = \frac{\pi}{2} + n \pi = \frac{\left(2 n + 1\right) \pi}{2}$, where $n$ is any integer.

Hence, the vertical asymptotes are

$x = \frac{\left(2 n + 1\right) \pi}{2}$, where $n$ is any integer.