Help me show that the limit $\lim_{(x,y)\to (0,0)}\frac{2e^x y^2}{x^2+y^2}$ does not exist 
Show that the limit $\lim_{(x,y)\to (0,0)}\frac{2e^x y^2}{x^2+y^2}$
  does not exist

$$\lim_{(x,y)\to (0,0)}\frac{2e^x y^2}{x^2+y^2}$$
Divide by $y^2$:
$$\lim_{(x,y)\to (0,0)}\frac{2e^x}{\frac{x^2}{y^2}+1}$$
$$=\frac{2(1)}{\frac{0}{0}+1}$$
Since $\frac{0}{0}$ is undefined. This limit does not exist.
I am not satisfied with my proof. Makes me think that what I just did was simple step, and not acceptable university level mathematics. Any comments on my proof?
 A: Indeed your proof is not valid.
Let $y=x$, $\lim_{(x,y)\to (0,0)}\frac{2e^x y^2}{x^2+y^2}=1$
Let $y=2x$, $\lim_{(x,y)\to (0,0)}\frac{2e^x y^2}{x^2+y^2}=\large{8\over5}$
So it cannot exist.
A: If we let $(x,y)\to(0,0)$ on the $x$ axis we have
$$f(x,0)=\frac{2e^x(0)^2}{x^2+(0)^2}=0\quad\implies \quad f(x,y)\to0\quad\text{as } (x,y)\to(0,0)\text{ on the }x\text{ axis}$$
On the $y$ axis we have
$$f(0,y)=\frac{2e^0y^2}{0^2+y^2}=2\quad\implies \quad f(x,y)\to 2\quad\text{as } (x,y)\to(0,0)\text{ on the }y\text{ axis}$$
A: The limit exists if for any $\varepsilon>0$ there exists a $\delta>0$ such that
$$\frac{2e^x y^2}{x^2+y^2}<\varepsilon\tag 1$$
if the distance of $(x,y)$ from the origin is less than $\delta$. That is, the limit exists only if it is independent of the path of $(x,y)$ when approaching $(0,0)$.
In the case of functions of one variable the parallel statement is that the limit exits if the  following two limits exist and equal
$$\lim_{x\downarrow 0}f(x)=\lim_{x\uparrow 0}f(x).$$

I am going to show that in the case of the limit at stake $(1)$ is not satisfied. 
So far so good:
$$\lim_{(x,y)\to (0,0)}\frac{2e^x y^2}{x^2+y^2}=\lim_{(x,y)\to (0,0)}\frac{2e^x}{\frac{x^2}{y^2}+1}.$$
At this point, however, let's say that $y=mx$, that is, $(x,y)$ approaches $(0,0) $along the line $y=mx$. Substituting this, we get
$$\lim_{(x,y)\to (0,0)}\frac{2e^x}{\frac{1}{m^2}+1}=\frac{2}{\frac1{m^2}+1}.$$
The limit does not exist because its value depends on the path we approach the origin.
A: In addition to the answers showing you a way: when dealing with limits you can't just put in the values for (in this case) $x$ and $y$. With the same argument you would say, that $$\lim\limits_{x\to 0} \frac{\sin(x)}{x}=\frac{0}{0}=\text{undefined}$$ whereas it is easy to show (and something one should know) that $$\lim\limits_{x\to 0} \frac{\sin(x)}{x}=1.$$ Just saying that $\frac{0}{0}$ is undefined is not enough, the limit might still exist. With more than one variable this becomes even more complicated as there are "more ways" to approach $(0,0)$.
You also might find this question helpful.
