# Where is my mistake in finding this lim ? $\lim\limits_{(x,y)\to (0,0)}\frac{|y|}{x^{2}}~e ^{-\frac{|y|}{x^{2}}}$

Im going to find the mistake in this two solutions :

Question $$\to$$ find :

$$\Omega =\lim\limits_{(x,y)\to (0,0)}\frac{|y|}{x^{2}}~e ^{-\frac{|y|}{x^{2}}}$$

The first suggested solution

Let prove does not exist lim :

$$•\color{red}{y=x^{2}}$$ then :

$$\Omega =\lim\limits_{(x,x^{2})\to (0,0)}\frac{|x^{2}|}{x^{2}}~e ^{-\frac{|x^{2}|}{x^{2}}}$$

$$=\lim\limits_{(x,x^{2})\to (0,0)}e ^{-\frac{x^{2}}{x^{2}}}=\color{green}{\frac{1}{e}}$$

$$•\color{red}{y=x}$$ then :

$$\Omega =\lim\limits_{(x,x)\to (0,0)}\frac{|x|}{x^{2}}~e ^{-\frac{|x|}{x^{2}}}$$

$$\Omega =\lim\limits_{(x,y)\to (0,0)}\frac{1}{|x|}~e ^{-\frac{1}{|x|}}$$ $$=\lim\limits_{t\to +\infty}te^{-t}=\color{green}{0}$$ This mean that : does not exist lim!

The second suggested solution

Using the polar coordinates, we find:

$$x=r\cos \theta , y=r\sin \theta$$

So :

$$\Omega =\lim\limits_{r\to 0}\frac{|\sin \theta |}{r\cos^{2} \theta }e^{-\frac{|\sin \theta |}{r\cos^{2} \theta }}$$

$$=\lim\limits_{t\to +\infty}te^{-t}=\color{red}{0}$$

I am waiting for your explanation, comments and advice, I will be happy if i see other ways!

Thanks!

It is wrong to say that $$\frac {|sin \theta | }{ r\cos^{2} \theta} \to \infty$$ as $$r \to 0$$. What happens when $$\sin \theta =0$$?
Assume you have to calculate $$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right)$$ and you want to use the trigonometrical substitution $$x=r\cos \theta$$ $$y=r \sin \theta$$ Assume that $$\mathop {\lim }\limits_{r \to 0} f\left( {r\cos \theta ,r\sin \theta } \right) = L$$ Then it is true that $$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = L$$ if and only if $$L$$ is independent from $$\theta$$.
Different paths can be interpreted in polar form via different angles through which we approach the limit. The function behaves differently when you approach the origin through two different lines $$\theta=0$$ and $$\theta=\pi/2$$ . Thus even in your polar approch also, conclusion would be the same.