Answer
= $0$
Work Step by Step
$\mathop {\lim }\limits_{n \to 0^{+} }$ $e^{\frac{-1}{y}}\ln{y}$
= $\mathop {\lim }\limits_{n \to 0^{+} }$ $\frac{\ln{y}}{e^{\frac{1}{y}}}$
= $\mathop {\lim }\limits_{n \to 0^{+} }$ $\frac{-y^{2}}{ye^{\frac{1}{y}}}$
= $\mathop {\lim }\limits_{n \to 0^{+} }$ $\frac{-y}{e^{\frac{1}{y}}}$
= $0$
