differentiate f(x) using L'hopital and other problem 
*

*Evaluate: $\ \ \ \ \lim_{x\to1}(2-x)^{\tan(\frac{\pi}{2}x)}$

*Show that  the  inequality holds: $\ \ \ x^\alpha\leq \alpha x + (1-\alpha)\ \ \ (x\geq0, \,0<\alpha <1)$
Please help me with these. 
Either a hint or a full proof will do.
Thanks.
 A: Just a good useful point for the first one. You can use this fact that  if $\lim_{x\to{+\infty}} f(x)^{g(x)}=1^{+\infty}$, which is indeterminate limit as we have here, so then we can solve it by  the following way instead:  

If $k =\lim_{x\to +\infty}\big(f(x)-1\big)g(x)$ then $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^k$$

A: For the second question, we consider the function $f(x) = x^\alpha - \alpha x$. We notice the $f'(x) = \alpha x^{\alpha-1} - \alpha$. $f'(1)=0$. We then look at the second derivative $f''(x)=\alpha(\alpha-1)x^{\alpha-2}<0$ for all $0<\alpha<1$ and $x>0$ This means $f$ is concave and 1 must be the global maximum. i.e. we have $x^\alpha - \alpha x\leq 1-\alpha$ and therfore $x^\alpha \leq \alpha x + 1-\alpha$
A: For the first problem: another useful method is to log the function to get $\frac{\log(2-x)}{\frac{1}{\tan(\frac{\pi x}{2})}}$, now you can apply L'Hospital's rule, take the limit and exponentiate it. I keep getting $e^{\frac{\pi}{2}}$, but you need to check the algebra more thoroughly
