How to apply Cauchy's MVT to evaluate the following? 
Use Cauchy's Mean Value theorem to evaluate 
$$\lim_{x\rightarrow 1} \left[\frac{\cos(\frac{1}{2}\pi
 x)}{\ln(1/x)}\right]$$

I can't understand how to apply Cauchy's MVT over here. Any hints? 

$$\lim_{x\to 1}\frac{\cos\left(\frac{\pi}{2}x\right)}{-\log x}\stackrel{x\mapsto 1-z}{=}\lim_{z\to 0}\frac{\sin\left(\frac{\pi}{2}z\right)}{-\log(1-z)}$$
and since $\lim_{z\to 0}\frac{\sin z}{z}=1=\lim_{z\to 0}\frac{z}{-\log(1-z)}$ the wanted limit is $\frac{\pi}{2}$, you do not need anything fancy. If you like, you may apply de l'Hopital's rule to get
$$ \lim_{x\to 1}\frac{\cos\left(\frac{\pi}{2}x\right)}{-\log x}=\frac{\pi}{2}\lim_{x\to 1}x\sin\left(\frac{\pi}{2}x\right)=\frac{\pi}{2}$$
or consider that
$$\lim_{x\to 1}\frac{\cos\left(\frac{\pi}{2}x\right)}{-\log x}\stackrel{x\mapsto e^z}{=}\frac{\pi}{2}\lim_{z\to 0}\frac{1}{z}\int_{0}^{z}e^t\sin\left(\frac{\pi}{2}e^t\right)\,dt=\frac{\pi}{2}\lim_{z\to 0}\int_{0}^{1}e^{zt}\sin\left(\frac{\pi}{2}e^{zt}\right)\,dt $$
and reach the same conclusion through the dominated convergence theorem.
