Finding the limit of $x \sin\frac{\pi}{x}$ How can I find the limit of the following
$x\rightarrow\infty$
$x \sin\frac{\pi}{x}$
I did 
$\dfrac{\sin\frac{\pi}{x}}{\frac{1}{x}}$
I took the derivative using l hospital and got.
$\dfrac{-1x^{-2} \cos \dfrac{\pi}{x}}{-1x^{-2}}$
Simplying I get
$\cos \frac{\pi}{x}$ but I am stuck.
another problem I have is
$\dfrac{\ln(x)}{\cot x}$ as $x\rightarrow0$
I did 
$\dfrac{\dfrac{1}{x}}{-\csc^2(x)}$
But I am unsure how to go on.
 A: Ok, so now you want to evaluate
$$
\lim_{x\to\infty} \cos\left(\frac{\pi}{x}\right)
$$
what is $\lim_{x\to\infty} \dfrac{\pi}{x}$? What do you get if you plug that into $\cos$?
For the second limit, you can rewrite it as
$$
-\dfrac{\sin^2(x)}{x} = -\sin(x) \dfrac{\sin(x)}{x}
$$
You should know what $\lim_{x\to 0}\dfrac{\sin(x)}{x}$ is, and then use 'limit of product is product of limits'.
A: Simply introduce a new variable $t = 1/x$, then the problem becomes $\lim_{t\rightarrow0}\dfrac{\sin\pi t}t$ which you know the answer ($\pi$).
A: For problems like this you can expand sine in terms of the following taylor series:
$$ sin (p) = p - \frac{p^3}{3!} + \frac{p^5}{5!} - \frac{p^7}{7!} + ...$$
Substituting $p = \frac{\pi}{x}$, we get:
$sin(\frac{\pi}{x}) = \frac{\pi}{x} - \frac{\pi^3}{x^3 3!} + \frac{\pi^5}{x^5 3!} - ...$
We are evaluating $f(x) = x \space sin(\frac{\pi}{x})$, so if we multiply above by $x$, we see that the first term is $\pi$, the other terms are powers of $\frac{1}{x}$
Hence when you apply the limit, the answer tends to $\pi$
