Answer
$0$
Work Step by Step
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, $\lim\limits_{x \to (\frac{\pi}{2})^{+}}e^{tanx}=0$
