<s>[INST] Several (advanced) L'Hospital problems Problems : 
$$
\begin{align}
&\text{A}.\ \lim_{x\rightarrow1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{B}.\ \lim_{x\rightarrow 0}\left(\cot x-\frac{1}{x}\right)\\
&\text{C}.\ \lim_{x\rightarrow0}\frac{\sin^{-1}x-x}{\tan^{-1}x-x}\\
&\text{D}.\ \lim_{x\rightarrow 1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{E}.\ \lim_{x\rightarrow0}\left(a^x+b^x\right)^\frac{1}{x}
\end{align}
$$

Here are my solutions concerning A~E.
A. $$\ln y=\tan\left(\frac{\pi x}{2}\right)\ln(x-2)=\frac{\tan\left(\frac{\pi x}{2}\right)}{\frac{1}{\ln(2-x)}}$$
$$\lim_{x\rightarrow 1}\frac{\tan\left(\frac{\pi x}{2}\right)}{\frac{1}{\ln(2-x)}}=_H\lim_{x\rightarrow1}\frac{\frac{\pi}{2}\sec^2(\frac{\pi x}{2})}{x-2}=\infty$$so, $$A\rightarrow \infty$$
Is this conclusion right?
C.
$$=_H\lim_{x\rightarrow0}\frac{\frac{1}{\sqrt{1-x^2}}-1}{\frac{1}{1+x^2}-1}=_H\lim_{x\rightarrow0}\frac{\frac{-1}{2}\frac{1}{\sqrt{(1-x^2)^3}}(-2x)}{-\frac{2x}{(1+x^2)^2}}=\frac{1}{-2}=-\frac{1}{2}$$
 [/INST] For example:
$$\lim_{x\to 0}\cot x-\frac1x=\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin x}\stackrel{\text{l'Hospital}}=\lim_{x\to 0}\frac{\cos x-x\sin x-\cos x}{\sin x+x\cos x}=$$
$$=-\lim_{x\to 0}\frac{\sin x}{\frac{\sin x}x+\cos x}=-\frac0{1+1}=0$$
</s>