Answer
Does not exist
Work Step by Step
Evaluate the limit along the curve $y=x^2$:
$\lim\limits_{(x,y) \to (0,0)}\frac{x^2ye^y}{x^4+4y^2}=\lim\limits_{x \to0}\frac{x^2\cdot x^2e^{x^2}}{x^4+4(x^2)^2}=\lim\limits_{x\to 0}\frac{x^4e^{x^2}}{5x^4}=\lim\limits_{x \to 0}\frac{e^{x^2}}{5}=\frac{e^0}{5}=\frac{1}{5}$
Evaluate the limit along the curve $y=-x^2$:
$\lim\limits_{(x,y) \to (0,0)}\frac{x^2ye^y}{x^4+4y^2}=\lim\limits_{x \to0}\frac{x^2\cdot (-x^2)e^{x^2}}{x^4+4(-x^2)^2}=\lim\limits_{x\to 0}\frac{-x^4e^{x^2}}{5x^4}=\lim\limits_{x \to 0}\frac{-e^{x^2}}{5}=\frac{-e^0}{5}=\frac{-1}{5}$
Since the values of the limit along two curves $y=x^2$ and $y=-x^2$ which lie $(0,0)$ are different, it concludes that the value of the limit does not exist.
