Limit evaluation $\lim_{x\to \pi/2}\cos(x)^{2x-\pi}$ I need to solve the following limit:
$$ \lim_{x\to \pi/2}\cos(x)^{2x-\pi} $$
I attempted to use natural logarithm:
$$
\lim_{x\to \pi/2} (2x-\pi)(\ln(\cos x))=
$$
$$
\lim_{x\to\pi/2} (2x\ln(\cos x)-\pi\ln(\cos x))=
$$
$$
2\times\lim_{x\to \pi/2} \frac{\ln(\cos x)}{x^{-1}}-\pi\times\lim_{x\to \pi/2}\ln(\cos x))=
$$
Applying LH twice to the first term gives
$$
2\times\lim_{x\to \pi/2} \frac{2x\cos x}{-\sin x}-\pi\times\lim_{x\to \pi/2}\ln(\cos x))=0-\pi\infty=-\infty
$$
And $e^{-\infty}=0$
However, for some reason, the correct result is $1$. Where's the flaw in my approach and what's the correct solution?
 A: That is not an indefinite form, L'Hospital doesn't apply.
By substitution $y=\pi/2-x$, we are to compute $\lim_{y\rightarrow 0^{+}}\left(\cos(\pi/2-y)\right)^{-2y}=\lim_{y\rightarrow 0^{+}}(\sin y)^{-2y}$.
While $\log(\sin y)^{y}=y\log((\sin y/y)\cdot y)=y\log(\sin y/y)+y\log y\rightarrow 0\cdot\log 1+0=0$, so $(\sin y)^{y}\rightarrow 1$.
Note that $y\log y\rightarrow 0$ can be done by L'Hopital or $|y\log y|=y\log(1/y)\leq Cy\cdot(1/y)^{1/2}$ for small $y>0$.
A: Here is a slick way to interpret this:
\begin{eqnarray*}
\lim_{x\to\frac{\pi}{2}} \cos(x)^{2x - \pi} & = & \lim_{x\to \frac{\pi}{2}}\operatorname{Re}\left[\left (e^{ix}\right )^{2x-\pi} \right ] \\
& =&\operatorname{Re}\left [\lim_{x\to \frac{\pi}{2}} e^{2ix^2 - i\pi x}\right ] \\
& = &\operatorname{Re}\left [ e^{2i\frac{\pi^2}{4} - i\frac{\pi^2}{2}} \right ] \\
& = &\operatorname{Re}\left [e^0\right ] \\
& = & 1.
\end{eqnarray*}
A: I think that answer is 1. You can use the the second remarkable limit lim(1+x)^(1/x) x->0
then we can convert function so lim(1+(cosx-1))^(1/(cos(x)-1))(cosx-1)(2x-pi)=
e^(lim (cos(x)-1)*(2x-pi)) x->pi/2 => lim (-sin^2(x))(2x-pi)/(cosx+1) x->pi/2 => e^0=1 
A: Note that we need to consider $x\to \frac \pi 2^-$ in order to have $\cos x\to 0^+$.
That’s a nice approach, from here we can use that
$$(2x-\pi)(\ln(\cos x))=  \frac{2x-\pi}{\cos x}(\cos x)\ln(\cos x)\to 0$$
indeed by standard limits
$$\cos x\ln(\cos x)\to 0$$
$$\frac{2x-\pi}{\cos x}= -2\frac{\frac{\pi}2-x}{\sin\left(\frac{\pi}2-x\right)}\to -2$$
A: First, observe this limit can make sense only if $x\nearrow \frac\pi2$, because $\cos x $ has to be positive. So, set $u=\frac\pi2-x\enspace (u\searrow 0)$.
Finding the limit amounts to finding the limit of the log:
$$ (2x-\pi)\ln(\cos x)=-2u\ln\Bigl(\cos\bigl(\tfrac\pi 2-u\bigr)\Bigr)=-2u\ln(\sin u). $$
Now, $\sin u\sim_0 u$, so $\;-2u\ln(\sin u)\sim_{u\to0^+}-2u\ln u\to 0 $, by a standard high-school limit. Therefore, by continuity og the exponential function, the required limit is equal to $1$.
