Limits calculus very short question? Can you help me to solve this limit? $\frac{\cos x}{(1-\sin x)^{2/3}}$... as $x \rightarrow \pi/2$, how can I transform this?

To evaluate the limit:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
we can use the fact that as $x$ approaches $\frac{\pi}{2}$, $\sin x$ approaches 1 and $\cos x$ approaches 0. Substituting these values into the expression, we get:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{\left(1-\sin x\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{\left(1-1\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{0^{\frac{2}{3}}}\right)$$
The expression is indeterminate in this form ($\frac{0}{0}$), so we can try using L'Hopital's rule to evaluate the limit. Differentiating the numerator and denominator with respect to $x$ gives us:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-3\left(1-\sin x\right)^{-\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-3\left(1-\sin x\right)^{-\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\cos\left(x\right)}{3\left(-\frac{1}{3}\right)\left(1-\sin x\right)^{-\frac{4}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{-1\left(1-\sin x\right)^{\frac{4}{3}}}\right)$$
Again, the expression is indeterminate in this form ($\frac{0}{0}$), so we can try using L'Hopital's rule again. Differentiating the numerator and denominator with respect to $x$ gives us:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-4\left(1-\sin x\right)^{\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-4\left(1-\sin x\right)^{\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\cos\left(x\right)}{4\left(\frac{1}{3}\right)\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
$$= \lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{1\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
Substituting the values $\sin x = 1$ and $\cos x = 0$, we get:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{0}{\left(1-1\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{0^{\frac{2}{3}}}\right)$$
The expression is indeterminate in this form ($\frac{0}{0}$), so we can't use L'Hopital's rule any further. However, since $\cos x$ approaches 0 and $\left(1 - \sin x\right)^{\frac{2}{3}}$ approaches 0 as $x$ approaches $\frac{\pi}{2}$, the entire expression approaches $\frac{0}{0}$, which is indeterminate. Therefore, the limit does not exist
