# How do you find vertical asymptotes in calculus?

Aug 17, 2015

Vertical asymptotes tend to be found whenever an x-intercept cannot be found for individual $x$ values. Sometimes you just have to understand the domain of a particular function to realize where these asymptotes would be, or you can solve for them.

If you have:

${x}^{2} / \left(\left(x - 2\right) \left(x + 3\right)\right)$

then I would expect asymptotes at $x = 2$ and $x = - 3$, like so:

graph{x^2/((x-2)(x+3)) [-10, 10, -5, 5]}

This one has no limit at those values because the limit from each side is different than the one from the other side (e.g. $- \infty$ vs. $\infty$).

As an example of solving for one, set $y = 0$ and solve for $x$, and you should be able to find the x-intercepts that exist.

$0 = {x}^{2} / \left(\left(x - 2\right) \left(x + 3\right)\right)$

Since you get $0 = {x}^{2}$ overall, an x-intercept exists at $x = 0$, but since you had to multiply $0 \cdot \left(x - 2\right) \left(x + 3\right)$, neither the $x = 2$ nor the $x = - 3$ intercepts exist.

If you have:

$\tan x$

then I would expect asymptotes at $\frac{\pi}{2} \pm \pi k$ where $k$ is an integer, like so:

graph{tanx [-10, 10, -5, 5]}

Same thing with regards to limits from either side of each vertical asymptote.

$0 = \tan x = \frac{\sin x}{\cos x}$

Since $0 = \sin x$, an intercept exists at $x = \pm \pi k$ where $k$ is an integer, but since you had to perform $0 \cdot \cos x$, asymptotes exist wherever $\cos x = 0$, namely, $x = \frac{\pi}{2} \pm \pi k$ where $k$ is an integer. Hence, $\tan x$ has those asymptotes.