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

Apr 30, 2017

$x = \frac{8}{k}$ where $k \in \mathbb{Z}$

#### Explanation:

Vertical asymptotes occur when the denominator equals $0$.

Focusing on the non-trigonometric part first:

$2 {x}^{2} - 8 = 0$

${x}^{2} = 4$

$x = \pm 2$

The other part is more difficult:

$\cos \left(\frac{\pi}{2} + \frac{8 \pi}{x}\right) = 0$

$\cos \left(x\right) = 0$ for $x = \frac{\pi}{2} , \frac{3 \pi}{2} , \frac{5 \pi}{2}$, which can be expressed as $x = \frac{\pi}{2} + k \pi$, where $k \in \mathbb{Z}$ (which means that $k$ is an integer).

This means that $\cos \left(\frac{\pi}{2} + \frac{8 \pi}{x}\right) = 0$ and $f \left(x\right)$ is asymptotic when:

$\frac{\pi}{2} + \frac{8 \pi}{x} = \frac{\pi}{2} + k \pi \text{ "" "" } , k \in \mathbb{Z}$

So:

$\frac{8 \pi}{x} = k \pi \text{ "" "" } , k \in \mathbb{Z}$

$\frac{8}{x} = k \text{ "" "" } , k \in \mathbb{Z}$

$x = \frac{8}{k} \text{ "" "" } , k \in \mathbb{Z}$

So there are vertical asymptotes at $x = \pm \frac{8}{1} , \pm \frac{8}{2} , \pm \frac{8}{3} , \pm \frac{8}{5} , \pm \frac{8}{6.} . .$

Note that the values found from the $2 {x}^{2} - 8$ portion are included in this set, as a sort of special case where $k = \pm 4$.