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!



3 Answers 3


The first suggested solution is correct.

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.


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