Answer
$-2$
Work Step by Step
We will calculate the limit for $\lim\limits_{(x,y) \to (\pi,\ln 2)} e^y \cos x$
This implies that $\lim\limits_{(x,y) \to (\pi,\ln 2)} e^y \cos x=e^{(\ln 2)} \cos (\pi)$
Thus, $\lim\limits_{(x,y) \to (\pi,\ln 2)} e^y \cos x=e^{(\ln 2)} \cos (\pi)=2 \times (-1)=-2$
