# Where are the vertical asymptotes of f(x) = cot x?

Jan 27, 2015

Vertical asymptotes are related to the domain of a function, and in particular to the point excluded from the domain because they cause a denominator to be 0.

Although it's quite unproperly written, indeed, it has a sense to write that $\frac{a}{0} = \setminus \pm \setminus \infty$, depending on the positivity of $a$.

The point is that you cannot divide by zero, so you must interpret the "division by zero" as a limit process, in which you divide by smaller and smaller quantity; and you'll find out that (if the denominator is fixed) as the denominator gets smaller, the absolute value of the fraction gets bigger and bigger.

This process is visually shown with vertical asymptotes, since you cannot evaluate the function in the fobidden point, but you can evaluate the function in points which lay as near as you want to the exlcluded one. And for what we just said, the value of the function will get bigger and bigger (positively or negatively) as you approach the zero of the denominator.

In your case, the function $\setminus \cot \left(x\right)$ is defined as $\frac{1}{\setminus \tan \left(x\right)}$, which is $\frac{\cos \left(x\right)}{\setminus \sin \left(x\right)}$. So, the zeros of the denominator are the ones of the sine function which, periodicity apart, are $0$ and $\setminus \pi$.

So, your vertical asymptotes are vertical lines of equations $x = 0$ and $x = \setminus \pi$. Considering periodicity, all the vertical asymptotes of $\setminus \cot \left(x\right)$ are of the form $x = k \setminus \pi$, for some $k \setminus \in \setminus m a t h \boldsymbol{Z}$.