<s>[INST] Differentiation using l´Hopital I need to use L´Hopital's rule with this functions:
$$\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}$$
$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$$
I take the exponent down using the properties of logarithms and then make the denominator like: $\lim_{x\rightarrow\frac{\pi}{2}} \frac{\cos(x)}{\frac{1}{\ln(1-\sin(x))}}$ but I still get stuck.
 [/INST] For the first one we have
$$(1-\sin x)^{\cos x}= (1-\sin x)^{\cos x}\frac{(1+\sin x)^{\cos x}}{(1+\sin x)^{\cos x}}=\frac{(\cos^2 x)^{\cos x}}{(1+\sin x)^{\cos x}}\to \frac 11=1$$
indeed by standard limits


*

*$\lim_{x\rightarrow\frac{\pi}{2}} (\cos^2 x)^{\cos x}=\lim_{t\to 0}(t^2)^t=1$

*$\lim_{x\rightarrow\frac{\pi}{2}} (1+\sin x)^{\cos x}=2^0=1$
For the second one
$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan x)^{\tan(2x)}=\lim_{x\rightarrow\frac{\pi}{4}} \left[(1+(\tan x-1))^\frac{1}{\tan x-1}\right]^{\tan(2x)(\tan x-1)}\to e^{-1}=\frac1e$$
indeed by standard limits


*

*$\lim_{x\rightarrow\frac{\pi}{4}} (1+(\tan x-1))^\frac{1}{\tan x-1}\to e$

*$\lim_{x\rightarrow \frac{\pi}{4}} \tan(2x)(\tan x-1)=\lim_{x\rightarrow \frac{\pi}{4}} \frac{2\tan x}{1-\tan^2 x}(\tan x-1)=\lim_{x\rightarrow \frac{\pi}{4}} \frac{-2\tan x}{1+\tan x}=-1$
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