Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{y + x{e^{ - {y^2}}}}}{{1 + {x^2}}} \cr
& {\text{Evaluate by direct substitution, 0 for }}x{\text{ and }}0{\text{ for }}y \cr
& = \frac{{0 + \left( 0 \right){e^{ - {{\left( 0 \right)}^2}}}}}{{1 + {{\left( 0 \right)}^2}}} \cr
& = \frac{0}{1} \cr
& = 0 \cr} $$
