# Question about a particular limit (at infinity)

I have a question about limits, a problem specifically. I have been asked to solve the following limit in any way I see fit:

$$\lim_{x\to 2\pi^-}x\csc x$$

I know that the domain of $\csc$ is all numbers except for $n\pi$, and I know I could probably plug in numbers close to $2\pi$ to get the limit, and I'm pretty sure this is going to be a limit at infinity, but is there an easier way?

I feel like I should be able to do some limit-algebra and solve it that way, will I just have to look at the graph or plug in numbers? I was going to take out on of the x's and put it in front of the limit but because it is a variable I don't think I can do that. Any tips?

Thanks!

Hint: The limit is $$\lim_{x \rightarrow {2\pi}^{-}}\frac{x}{\sin x}$$ For values of $x$ close to but less than $2\pi,$ the values will be $$\frac{\text{numbers close to} \;\; 2\pi}{\text{negative numbers close to} \;\; 0}$$ which tells us the limit is $\ldots$
(Note that $\sin x$ is negative when $x$ is in the $4$th quadrant.)
Hint: $\csc(x)=\dfrac1{\sin(x)}$ and $\sin(2\pi)=0$.
Another Approach: Since $\sin(2\pi-\epsilon)=-\sin(\epsilon)$ and for $0\le\epsilon\le\pi$, we have $0\le\sin(\epsilon)\le \epsilon$ you can use $x=2\pi-\epsilon$ to show that $$\lim_{\epsilon\to0^+}\frac{2\pi-\epsilon}{-\sin(\epsilon)}\le-\lim_{\epsilon\to0^+}\frac\pi\epsilon$$