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?
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.
$\displaystyle \lim_{x\rightarrow\pi/2}\frac{\cos(x)}{(1-\sin(x))^{2/3}}=\displaystyle \lim_{x\rightarrow\pi/2} \frac{\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}}{(\sin^2{\frac{x}{2}}-2\sin{\frac{x}{2}}\cdot \cos {\frac{x}{2}}+\cos^2{\frac{x}{2}})^{2/3}}=$

$=\displaystyle \lim_{x\rightarrow\pi/2} \frac{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})(\cos {\frac{x}{2}}+\sin {\frac{x}{2}})}{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})^{4/3}}=\displaystyle \lim_{x\rightarrow\pi/2} \frac {\cos {\frac{x}{2}}+\sin {\frac{x}{2}}}{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})^{1/3}}=\frac{\sqrt{2}}{0}=\infty$