Answer
$${e^{ - 7}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1, - 3} \right)} {e^{2x - {y^2}}} \cr
& {\text{Using the limit laws }} \cr
& = {e^{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1, - 3} \right)} \left[ {2x - {y^2}} \right]}} \cr
& {\text{then}}{\text{, substituting }}1{\text{ for }}x{\text{ and }} - 3{\text{ for }}y \cr
& = {e^{2\left( 1 \right) - {{\left( 3 \right)}^2}}} \cr
& {\text{simplifying}} \cr
& = {e^{2 - 9}} \cr
& = {e^{ - 7}} \cr} $$
