Answer
$$\lim_{\theta\to0^+}\sqrt{\theta}e^{\cos(\pi/\theta)}=0$$
Work Step by Step
$$A=\lim_{\theta\to0^+}\sqrt{\theta}e^{\cos(\pi/\theta)}$$
We know that $-1\le \cos(\pi/\theta)\le1$
So $-\sqrt\theta e^{-1}\le\sqrt{\theta}e^{\cos(\pi/\theta)}\le-\sqrt\theta e^{1}$
This equals to, $-\sqrt\theta e^{-1}\le\sqrt{\theta}e^{\cos(\pi/\theta)}\le-\sqrt\theta e$
- $\lim_{\theta\to0^+}(-\sqrt\theta e^{-1})=\sqrt0 e^{-1}=0$
- $\lim_{\theta\to0^+}(-\sqrt\theta e)=\sqrt0 e=0$
Therefore, according to the Sandwich Theorem, we can conclude that
$$A=\lim_{\theta\to0^+}\sqrt{\theta}e^{\cos(\pi/\theta)}=0$$
