## Calculus 8th Edition

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Need to find the limit for $\lim\limits_{x \to (\frac{\pi}{2})^{+}}e^{tanx}$. Since, $tan(\frac{\pi}{2})^{+}=-\infty$ Thus, $\lim\limits_{x \to (\frac{\pi}{2})^{+}}e^{tanx}=e^{-\infty}=\frac{1}{e^{\infty}}$ Hence, $\lim\limits_{x \to (\frac{\pi}{2})^{+}}e^{tanx}=0$