Limit $\frac{0}{0}$ which tends to $\frac{\pi}{2}$ I'm trying to evaluate the following limit:
$$\lim_{x\rightarrow\pi/2}\frac{\cos(x)}{(1-\sin(x))^{2/3}}$$
The limit has the form $\frac{0}{0}$, I've tried using L'Hopital's rule but I can't resolve it.
Any idea? 
 A: For a simple approach, we can use the following identities:
$1 - \sin 2x = (\cos x - \sin x)^2$
$\cos 2x = \cos ^2 x - \sin ^ 2 x$
Be careful when you consider the limits on either side.
A: As $x\to\pi/2$, $\cos(x) \sim (\pi/2 - x)$.  We also have $1 - \sin(x) = 1 - \cos(\pi/2 - x) \sim (1/2)(x - \pi/2)^2.$  Hence,
$${\cos(x)\over (1 - \sin(x))^{2/3}}\sim {\pi/2 - x\over(1/2)^{2/3} (x - \pi/2)^{4/3}}, $$
which blows up as $x\to\pi/2$.
A: Multiply top and bottom by $(1+\sin x)^{2/3}$:
$$\frac{\cos x}{(1-\sin x)^{2/3}}= \frac{\cos x(1+\sin x)^{2/3}}{(1-\sin x)^{2/3}(1+\sin x)^{2/3}}=\frac{\cos x(1+\sin x)^{2/3}}{(1-\sin^2 x)^{2/3}}=\frac{\cos x(1+\sin x)^{2/3}}{\cos^{4/3}x}.$$
Do the obvious cancellation. So we are interested in the behaviour of 
$$\frac{(1+\sin x)^{2/3}}{\cos^{1/3} x}$$
near $\pi/2$.  For $x$ close to $\pi/2$, but less than $\pi/2$, this is large positive, while on the other side of $\pi/2$ it is large negative.
Thus our limit does not exist.
Methodological comment: Taylor expansion, probably preceded by the change of variable $w=\pi/2-x$, remains the mathematically more natural approach. We are interested in local behaviour, and for nice functions that is precisely what the Taylor series gives us.
Pedantic comment: It is probably best to avoid non-integer powers of negative numbers. For one thing, the usual formal definition in terms of the logarithm breaks down. And when we deal with powers of negative numbers, familiar "rules" need to be used with caution.  
So instead of $\frac{\cos x(1+\sin x)^{2/3}}{\cos^{4/3}x}$, it would have been better to write $\frac{\cos x(1+\sin x)^{2/3}}{|\cos x|^{4/3}}$.  Then we can observe that this simplifies to $\frac{(1+\sin x)^{2/3}}{|\cos x|^{1/3}}$ if $\cos x$ is positive and to $\frac{-(1+\sin x)^{2/3}}{|\cos x|^{1/3}}$ if $\cos x$ is negative.
A: \begin{equation}
\lim_{x\to \pi/2}\frac{\cos x}{(1-\sin x)^{2/3}}=\lim_{x\to \pi/2}\frac{\sin(\pi/2-x)}{(1-\cos(\pi/2-x))^{2/3}}=\lim_{t\to 0}\frac{\sin t}{(1-\cos t)^{2/3}}=\lim_{t\to 0}\frac{t+o(t)}{(t^2+o(t^2))^{2/3}}
\end{equation}
The limit is $\infty$.
A: $\displaystyle \lim_{x\rightarrow\pi/2}\frac{\cos(x)}{(1-\sin(x))^{2/3}}=\displaystyle \lim_{x\rightarrow\pi/2} \frac{\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}}{(\sin^2{\frac{x}{2}}-2\sin{\frac{x}{2}}\cdot \cos {\frac{x}{2}}+\cos^2{\frac{x}{2}})^{2/3}}=$
$=\displaystyle \lim_{x\rightarrow\pi/2} \frac{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})(\cos {\frac{x}{2}}+\sin {\frac{x}{2}})}{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})^{4/3}}=\displaystyle \lim_{x\rightarrow\pi/2} \frac {\cos {\frac{x}{2}}+\sin {\frac{x}{2}}}{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})^{1/3}}=\frac{\sqrt{2}}{0}=\infty$
