Answer
\[\lim_{(x,y)\rightarrow (3,2)}e^{\sqrt{2x-y}}=e^2\]
Work Step by Step
Let \[l=\lim_{(x,y)\rightarrow (3,2)}e^{\sqrt{2x-y}}\]
\[\Rightarrow l=e^{\sqrt{2(3)-2}}\]
\[\Rightarrow l=e^{\sqrt{6-2}}\]
\[\Rightarrow l=e^{\sqrt{4}}\]
\[\Rightarrow l=e^{2}\]
Which is finite
So limit \[\lim_{(x,y)\rightarrow (3,2)}e^{\sqrt{2x-y}}\]
exists and
\[\lim_{(x,y)\rightarrow (3,2)}e^{\sqrt{2x-y}}=e^2\]
Hence , \[\lim_{(x,y)\rightarrow (3,2)}e^{\sqrt{2x-y}}=e^2\]
