Answer
$0$
Work Step by Step
Consider $f(y)=\lim\limits_{y \to 0^{+}}e^{-1/y} \ln y$
Let $\lim\limits_{y \to 0^{+}} \dfrac{\ln y}{e^{1/y}}=\lim\limits_{y \to 0^{+}} \dfrac{a(x)}{b(x)}$
Thus, $f(0)=\dfrac{\infty}{\infty}$
This shows an Inderminate form of the limit, so apply L'Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{y \to 0^{+}}\dfrac{1/y}{(-1/y^2)e^{1/y}}=\lim\limits_{y \to 0^{+}}\dfrac{-y}{e^{1/y}}=0$
