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.
 A: 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


*

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

*$\lim_{x\rightarrow \frac{\pi}{4}} \tan(2x)(\tan x-1)=\lim_{x\rightarrow \frac{\pi}{4}} \frac{2\tan x}{1-\tan^2 x}(\tan x-1)=\lim_{x\rightarrow \frac{\pi}{4}} \frac{-2\tan x}{1+\tan x}=-1$
A: $$\begin{align}
\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\ln(1-\sin(x))^{\cos(x)}\bigg)\\
&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\cos(x)\ln(1-\sin(x))\bigg)\\
&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{\ln(1-\sin(x))}{1/\cos(x)}\bigg)\\
&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{\frac{-\cos(x)}{1-\sin(x)}}{\frac{\sin(x)}{\cos^2(x)}}\bigg)\\
&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos^3(x)}{(\sin(x)-1)\sin(x)}\bigg)\\
&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos^3(x)}{\sin(x)-1}\bigg)\\
&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}\frac{-3\cos^2(x)\sin(x)}{\cos(x)}\bigg)\\
&=\exp\bigg(\lim_{x\rightarrow\frac{\pi}{2}}-3\cos(x)\sin(x)\bigg)\\
&=e^0\\
&=1
\end{align}$$
 I can tell you the answer for the second limit is $1/e$, but the technique is pretty much the same.
A: Without using L'Hospital:
$$\begin{align}1) \ \lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}&=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}\cdot 1=\\
&=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}\cdot \lim_{x\rightarrow\frac{\pi}{2}} (1+\sin(x))^{\cos(x)}=\\
&=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin^2(x))^{\cos(x)}=\\
&=\lim_{x\rightarrow\frac{\pi}{2}} (\cos^2(x))^{\cos(x)}\stackrel{(1)}=\\
&=1.\\
2) \ \lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}&=\lim_{x\rightarrow\frac{\pi}{4}} (1+\tan(x)-1)^{\frac{2\tan(x)}{(1-\tan x)(1+\tan x)}}=\\
&=\lim_{x\rightarrow\frac{\pi}{4}} (1+\tan(x)-1)^{-\frac{1}{\tan x-1}}\stackrel{(2)}=\\
&=e^{-1}.\end{align}$$
Note: 
$$\begin{align}\lim_\limits{x\to 0} x^x&=1 \quad (1)\\
\lim_\limits{x\to 0} (1+x)^{1/x}&=e \quad (2) \end{align}$$
A: Here is my take on the second and a remark on the first limit:
$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$: 
So, consider 
\begin{eqnarray*} \tan(2x)\ln(\tan x)
& \stackrel{\tan(2x)= \frac{2\tan x}{1-\tan^2x}}{=} & 2\frac{\tan x\cdot \ln(\tan x)}{1-\tan^2x}\\
& \stackrel{t = \tan x, t\to1}{=} & 2\frac{t \ln t}{1-t^2} \\
& \stackrel{L'Hosp.}{\sim} & 2\frac{\ln t + 1}{-2t}\\
& \stackrel{t\to 1}{\longrightarrow} & -1
\end{eqnarray*}
Hence, $\boxed{\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)} = e^{-1}}$.
$\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}$:
I recommend just to consider the logarithmic limit and not to write all time $e^{\lim ...}$. This is tedious and not necessary:
\begin{eqnarray*} \cos x \ln(1-\sin x)
& = & \frac{\ln(1-\sin x)}{\frac{1}{\cos x}}\\
& \stackrel{L'Hosp.}{\sim} & -\frac{\frac{\cos}{1- \sin x}}{-\frac{\sin x}{\cos^2 x}}\\
& = & \frac{\cos^3 x}{\sin x (1-\sin x)} \\
& = & \frac{\cos x (1-\sin^2 x)}{\sin x (1-\sin x)} \\
& \stackrel{\frac{1-\sin^2 x}{1-\sin x} = 1+\sin x}{=} & \frac{\cos x (1+\sin x)}{\sin x}\\
& \stackrel{x\to \frac{\pi}{2}}{\longrightarrow} & 0
\end{eqnarray*}
Hence, $\boxed{\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)} = e^0 = 1}$.
A: $$(1-\sin{x})^{\cos{x}}=e^{\cos{x}\log{(1-\sin{x})}}$$
$$\cos{x}\log{(1-\sin{x})}$$
$\cos{x} =\sqrt{1-\sin{x}}\sqrt{1+\sin{x}}$
So it suffices to find the limit $(\sqrt{1-\sin{x}})\log{(1-\sin{x})}$
Take $y=\sqrt{{1-\sin{x}}}$ so $y \to 0$ as $x \to \frac{\pi}{2}$
Thus $y\log{y^2} \to^{y \to 0} 0$ using L'Hospital's rule
and also $\sqrt{{1+\sin{x}}} \to \sqrt2$ as $x \to \frac{\pi}{2}$
So you can find the first limit.
