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? 
 A: $$\lim_{x\to 1}\frac{\cos\left(\frac{\pi}{2}x\right)}{-\log x}\stackrel{x\mapsto 1-z}{=}\lim_{z\to 0}\frac{\sin\left(\frac{\pi}{2}z\right)}{-\log(1-z)}$$
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.
A: The only way to apply Cauchy's MVT here that I can think of effectively amounts to applying L'Hôpital's rule (for this example, not in general). Since $\cos\frac{\pi}{2}=\ln1=0$, we can rewrite the given limit as:
$$\lim_{x\to1}\left[\frac{\cos\frac{\pi x}{2}}{-\ln x}\right]=\lim_{x\to1}\left[\frac{\cos\frac{\pi x}{2}-\cos\frac{\pi}{2}}{(-\ln x)-(-\ln1)}\right]=\lim_{x\to1}\left[\frac{f(x)-f(1)}{g(x)-g(1)}\right],$$
where $f(x)=\cos\frac{\pi x}{2}$ and $g(x)=-\ln x$. By Cauchy's MVT, for any $x\neq1$ there exists some $c$ between $1$ and $x$ ( i.e. $c\in[1,x]$ if $x>1$ or $c\in[x,1]$ if $x<1$) such that
$$\frac{\cos\frac{\pi x}{2}-\cos\frac{\pi}{2}}{(-\ln x)-(-\ln1)}=\frac{f(x)-f(1)}{g(x)-g(1)}=\frac{f'(c)}{g'(c)}=\frac{-\frac{\pi}{2}\sin\frac{\pi c}{2}}{-\frac{1}{c}}.$$
Since $c$ lies between $1$ and $x$, $x\to1$ forces $c\to1$, and so we end up with the limit of the last expression, which basically is (as I said above) an application of L'Hôpital's rule.
Maybe a totally different solution is expected, but I can't think of anything else …
