Solution to this limit

Question

Can anyone tell me how to calculate this limit. It is a puzzle so I think there must be some trick.

$lim_{x \to \dfrac{\pi}{20}} \left( {(\sec x)}^{\cos x} + {(\csc x)}^{\sin x} \right)$

Answer 1

$\sin$ and $\cos$ are analytic, so they're continuous everywhere, and in particular,

$$ \lim_{x \to \frac{\pi}{20}} \sin x = \sin \frac{\pi}{20} $$

and

$$ \lim_{x \to \frac{\pi}{20}} \cos x = \cos \frac{\pi}{20} $$

Since $\sin \frac{\pi}{20} \neq 0$ and $\cos \frac{\pi}{20} \neq 0$, the same applies to $\sec$ and $\csc$, so you can just evaluate this limit by substitution: replace all the $x$ with $\frac{\pi}{20}$ and you're done.

Answer 2

Notice, both the limits exist separately so we can break the given limit as follows $$\lim_{x\to \frac{\pi}{20}}\left(\sec^{\cos x}(x)+\csc^{\sin x}(x)\right)$$ $$=\lim_{x\to \frac{\pi}{20}}\left(\sec x\right)^{\cos x}+\lim_{x\to \frac{\pi}{20}}\left(\csc x\right)^{\sin x}$$ $$=\left(\sec \frac{\pi}{20}\right)^{\cos \frac{\pi}{20}}+\left(\csc \frac{\pi}{20}\right)^{\sin \frac{\pi}{20}}$$

Answer 3

$$\lim_{x \to \frac{\pi}{20}} \left( {(\sec(x))}^{\cos(x)} + {(\csc(x))}^{\sin(x)} \right)=$$ $$\lim_{x \to \frac{\pi}{20}} \sec^{\cos(x)}(x)+\lim_{x \to \frac{\pi}{20}} \csc^{\sin(x)}(x)=$$ $$\sec^{\cos\left(\frac{\pi}{20}\right)}\left(\frac{\pi}{20}\right)+\csc^{\sin\left(\frac{\pi}{20}\right)}\left(\frac{\pi}{20}\right)\approx 2.34901$$