# How do you find the vertical, horizontal and slant asymptotes of:  f(x) = x/ sinx?

Aug 6, 2016

$f \left(x\right)$ has vertical asymptotes at $x = n \pi$ for any non-zero integer $n$.

It has a hole at $x = 0$

#### Explanation:

Vertical asymptotes will occur where the denominator is zero and the numerator non-zero.

$\sin x = 0$ if and only if $x = n \pi$ for some $n \in \mathbb{Z}$

Hence $f \left(x\right)$ has vertical asymptotes at $x = n \pi$ where $n \in \mathbb{Z}$ and $n \ne 0$

$f \left(x\right)$ has a hole at $x = 0$. The rational expression becomes $\frac{0}{0}$, which is undefined, but the right and left limits exist at $x = 0$ and are equal:

${\lim}_{x \to 0} \frac{x}{\sin x} = 1$

$f \left(x\right)$ has no horizontal or slant asymptotes, since we find:

$f \left(2 n \pi + \frac{\pi}{2}\right) = 2 n \pi + \frac{\pi}{2} \to + \infty$ as $n \to + \infty$

$f \left(2 n \pi + \frac{\pi}{2}\right) = 2 n \pi + \frac{\pi}{2} \to - \infty$ as $n \to - \infty$

$f \left(2 n \pi + \frac{3 \pi}{2}\right) = - \left(2 n \pi + \frac{3 \pi}{2}\right) \to - \infty$ as $n \to + \infty$

$f \left(2 n \pi + \frac{3 \pi}{2}\right) = - \left(2 n \pi + \frac{3 \pi}{2}\right) \to + \infty$ as $n \to - \infty$