# For what values of x, if any, does f(x) = 1/x-tanx  have vertical asymptotes?

Dec 21, 2016

Vertical asymptotes occur at $x = 0 , \frac{\left(2 n + 1\right) \pi}{2}$ where $n \in \mathbb{N}$

#### Explanation:

We have

$f \left(x\right) = \frac{1}{x} - \tan x$

which we can write as;

$f \left(x\right) = \frac{1}{x} - \sin \frac{x}{\cos} x$

There will be vertical asymptotes when any part of a denominator is zero, so in this case:

For $\frac{1}{x}$ when $x = 0$
For $\sin \frac{x}{\cos} x$ when $\cos x = 0 \implies x = \frac{\pi}{2} , \frac{3 \pi}{2} , \frac{5 \pi}{2.} . .$

ie at $x = 0 , \frac{\left(2 n + 1\right) \pi}{2}$ where $n \in \mathbb{N}$

We can see these by looking at the graph: