# For what values of x, if any, does f(x) = 1/((x-4)sin(pi+(3pi)/x)  have vertical asymptotes?

Oct 11, 2017

A vertical asymptote occurs when $x = 4 , 0 , \mathmr{and} \frac{3}{n - 1} , n \in \mathbb{R}$

#### Explanation:

For there to be a vertical asymptote, the denominator bust be 0.

Here, we have $\left(x - 4\right) \sin \left(\setminus \pi + \frac{3 \setminus \pi}{x}\right)$ as the denominator. As long as one of these equals 0, the entire denominator will equal 0.

An asymptote will appear at $x = 4$ because if we put out $x$ value in, we get $0 \sin \left(\setminus \pi + \frac{3 \setminus \pi}{4}\right) = 0$.

A second asymptote will appear at $x = 0$ because in the $\sin$ we have a fraction with $x$ on the bottom, if this equals 0 then there can be no $y$ value.

The final asymptotes will appear when $x = \frac{3}{n - 1} , n \in \mathbb{R}$, this is because for $\sin \setminus \theta$ to equal 0, $\setminus \theta$ must equal $n \setminus \pi , n \in \mathbb{R}$.

So, we have $n \setminus \pi = \setminus \pi + \frac{3 \setminus \pi}{x}$

If we rearrange to make $x$ the subject, we get $x = \frac{3 \setminus \pi}{n \setminus \pi - \setminus \pi} = \frac{3}{n - 1}$

By putting this into $\sin \left(\setminus \pi + \frac{3 \setminus \pi}{x}\right)$, we get sin(\pi+(3\pi)/(3/(n-1)))-=sin(\pi+(3\pi(n-1))/3-=sin(\pi+\pi(n-1))-=sin(\pi+n\pi-\pi)-=sin(n\pi)