Find the vertical asymptotes (if any) of the function: f(x)=tan(-10x)
Determine on-sided limit: lim x-->6 from the left f(x)=csc(piX/6)
Any help is appreciated. Thanks
$\displaystyle \tan (-10x)=\frac{\sin(-10x)}{\cos (-10 x)}=-\frac{\sin 10 x}{\cos 10 x}.$
Points of discontinuity are the ones where you'll find vertical asymptotes. So you have to find those values x where $\displaystyle \cos 10 x = 0.$
That will be the case every time when $\displaystyle 10x=(2k-1)\cdot \frac{\pi}{2}$ for any $\displaystyle k\in \mathbb{Z}.$ So you see that this function has countably infinite number of points of discontinuity: $\displaystyle x=\frac{(2k-1)\pi}{20},\quad k\in\mathbb{Z}$ and for each of them you have one asymptote.
$\displaystyle \csc\left(\frac{\pi x}{6}\right)=$$\displaystyle \frac{1}{\sin\left(\frac{\pi x}{6}\right)}$
As x approaches 6 from the left, $\displaystyle \frac{\pi x}{6}$ approaches $\displaystyle \pi$ from the left. Of course, $\displaystyle sin(\pi)= 0$ so the only question is whether sin(x) is positive or negative for x slightly less than $\displaystyle \pi$.