<s>[INST] Evaluating limit of the function at $\frac{\pi}{2}$ I'm trying to solve this
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}$$
I have tried using the L'Hôpital's rule
But I'm stuck at
$$\lim_{x \to \frac{\pi}{2}} \frac{-\sin{x}}{3(x-\frac{\pi}{2})^2}$$
Since the above equation is not in the $\frac{0}{0}$ , $\frac{\infty}{\infty}$
or $\frac{anything}{\infty}$ form
Then I tried expanding the $\cos{x}$ as taylor series at $x=\frac{\pi}{2}$. Which on simplifying I am left with
$$ \lim_{x \to \frac{\pi}{2}} \frac{-1}{(x - \frac{\pi}{2})^2} + \frac{1}{6} - \frac{1}{120} (x - \frac{\pi}{2})^2 + \frac{(x - \frac{\pi}{2})^4}{5040} - ... $$
and I'm stuck again. How do I proceed ahead?
 [/INST] welcome to MSE
AS a hint
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}=\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi}{2})^3}$$now take $x-\frac{\pi}{2}=a $  when $x$ tends to $\frac{\pi}{2}$ ,a tends to zero
$$\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi}{2})^3}=\\
\lim_{a\to 0} \frac{\sin(-a)}{(a)^3}\\=
\lim_{a\to 0} \frac{-\sin(a)}{(a)^3}\\=
\lim_{a\to 0} \frac{-1}{(a)^2}\to -\infty$$
</s>