# For what values of x, if any, does f(x) = 1/((x-2)sin(pi+(8pi)/x)  have vertical asymptotes?

##### 1 Answer
Jun 7, 2016

This function has a vertical asymptote for $x = 2$ and for every other $x$ of the form $\frac{8}{k}$, for $k \setminus \in \setminus m a t h \boldsymbol{Z}$

#### Explanation:

A rational function has a vertical asymptote if its denominator equals zero. In your case, the denominator is

 (x-2)\sin(pi+(8pi)/x)

As every product, it equals zero if and only if at least one of its factors equals zero.

It is easy to find out that the first factor equals zero if $x = 2$.

As for the second, we can first of all note that $\sin \left(\setminus \pi + z\right) = - \setminus \sin \left(z\right)$

So, we need to find out when

 -\sin((8pi)/x)=0 \iff \sin((8pi)/x)=0

This function has an infinite number of solutions, given by

 (8pi)/x=k\pi \iff 8/x=k \iff x=8/k

So, for $x = \ldots - \frac{8}{4} , - \frac{8}{3} , - \frac{8}{2} , - 8 , 8 , \frac{8}{2} , \frac{8}{3} , \frac{8}{4.} . .$ the denominator equals zero, and the function has a vertical asymptote.