$\lim_{x \rightarrow \frac{\pi}{2}} \frac{(\frac{\pi}{2}-x)^2}{\cos x}$
1 Answer
We can L'hopital's rule
\[\lim_{x \rightarrow \frac{\pi}{2}} \frac{(\frac{\pi}{2}-x)^2}{\cos x}=\lim_{x \rightarrow \frac{\pi}{2}} \frac{-2(\frac{\pi}{2}-x)}{-\sin x}=\pm \infty.\]
Indeed the limit does not exists.
\[\lim_{x \rightarrow \frac{\pi}{2}} \frac{(\frac{\pi}{2}-x)^2}{\cos x}=\lim_{x \rightarrow \frac{\pi}{2}} \frac{-2(\frac{\pi}{2}-x)}{-\sin x}=\pm \infty.\]
Indeed the limit does not exists.
Erdos
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