Find the limit $ \cos x / (1 - \sin x)^{2/3}$ My problem:

Find the following limit 
  $$\lim\limits_{x \to \pi/2} \frac{\cos x}{\left(1 - \sin x\right )^{2/3}}$$

I tried $(1 - \sin x)^2 = 1 - 2\sin x + \sin^{2}x$ but can go as far as that.
 A: We have that by 


*

*$\cos x=\sin\left(\frac \pi 2 -x\right)$

*$\sin x=\cos\left(\frac \pi 2 -x\right)$
and $\frac \pi 2 -x =y \to 0$
$$\frac{\cos x}{\sqrt[3]{(1-\sin x)^2}} =\frac{\sin y}{\sqrt[3]{(1-\cos y)^2}}=\frac{\sin y}{y}\frac{\sqrt[3]{y^4}}{\sqrt[3]{(1-\cos y)^2}}\frac1{\sqrt[3]y} \to \pm \infty $$
indeed


*

*$\frac{\sin y}{y} \to 1$

*$\frac{\sqrt[3]{y^4}}{\sqrt[3]{(1-\cos y)^2}}=\sqrt[3]{\left(\frac{y^2}{{1-\cos y}}\right)^2}\to \sqrt[3] 2$

*$\frac1{\sqrt[3]y} \to \pm \infty$
A: Set $\dfrac\pi2-x=2y$
$$\lim_{y\to0^+}\dfrac{\sin2y}{(1-\cos2y)^{2/3}}=\lim_{y\to0^+}\dfrac{2\sin y\cos y}{2^{2/3}\sin^3y}=?$$
A: $${\cos x\over(1-\sin x)^{2/3}}={\cos x\over(1-\sin x)^{2/3}}\cdot{(1+\sin x)^{2/3}\over(1+\sin x)^{2/3}}={\cos x(1+\sin x)^{2/3}\over(1-\sin^2x)^{2/3}}={\cos x(1+\sin x)^{2/3}\over(\cos x)^{4/3}}={(1+\sin x)^{2/3}\over(\cos x)^{1/3}}\to{2^{2/3}\over\pm0}=\pm\infty\quad\text{as }x\to\pi/2$$
A: To make life easier, let $x=\frac \pi 2$ $$\lim\limits_{x \to \frac \pi 2} \frac{\cos (x)}{\left(1 - \sin(x)\right )^{2/3}}=-\lim\limits_{y \to 0}\frac{\sin(y)}{\left(1 - \cos(y)\right )^{2/3} }$$
Now, use the tangent half angle subsitution $y=2  \tan ^{-1}(t)$ to get 
$$\frac{\sin(y)}{\left(1 - \cos(y)\right )^{2/3} }=\frac{\sqrt[3]{2}}{\sqrt[3]{t^3+t}}\sim \frac{\sqrt[3]{2}}{\sqrt[3]{t}}$$ and $t \to 0^\pm$.
A: Empirical approach:
$\cos x$ will tend to $0$ linearly and $1-\sin x$ quadratically. Hence around $\frac\pi2$, the function behaves like the power $1-\frac23\cdot2$ of the gap, which diverges.

Rigorously:
$$\frac{\cos x}{(1-\sin x)^{2/3}}=\frac{\cos x}{(1-\sin^2x)^{2/3}}(1+\sin x)^{2/3}=\frac{(1+\sin x)^{2/3}}{\cos^{1/3}x}.$$
