About a limit-no l'hospital or derivatives So i've come across this limit:
$\lim_{x\to1} x^{\cot{\pi x}}$
So i found that its possible to solve without using those before-mentioned tools. 
I'm not sure how to do this limit,while i get stuck at this:
$\lim_{x\to 1} x^{\frac{\cos{\pi x} { \pi x}}{sin{\pi x}{  \pi x}}}$
I know i have to change the variable,so that i will $\lim_{t\to\infty}....$ for the modified variable $t$,  but i am stuck at that part.
Thank you in advance. Any help would be appreciated.
 A: Take $t=1-x$ and rewrite the expression as $ \exp( \cot(\pi (1-t)) \log (1-t) )$. The cosine just tends to $-1$, so it all amounts to evaluating $$ \lim_{t\to 0} \exp \left( - \frac{\log (1-t) }{\sin (\pi (1-t) )} \right) =  \lim_{t\to 0} \exp \left( - \frac{\log (1-t) }{t} \frac{\sin t}{\sin (\pi (1-t) )} \right) = e^{1/\pi}, $$ where we have used two elementary limits.
A: Apply $\ln$ to get
$$\tag 1\cot \pi x \ln x = \cos \pi x \frac{\ln x}{\sin \pi x}= \cos \pi x \frac{(\ln x-\ln 1)/(x-1)}{(\sin \pi x-\sin \pi\cdot 1)/(x-1)}.$$
As $x\to 1,\cos \pi x \to -1.$ In the fraction, use the definition of the derivative to see the top $\to \ln'(1) = 1,$ the bottom $\to$ the derivative of $\sin \pi x$ at $1,$ which is $\pi \cos \pi = - \pi.$ The limit of $(1)$ is thus $1/\pi.$ Exponentiating back gives $e^{1/\pi}$ for the original limit. (Note that the restriction $x\to 1^-$ is not needed. The full limit as $x\to 1$ is $e^{1/\pi}.$)
OK, I used the definition of the derivative and two basic derivative results. Are these not permitted here?
