# How to find $\lim_{(x,y)\rightarrow (0,0)} \frac{e^{-x^{2}} - y}{x^{2}+y^{2}}$?

$$\lim_{(x,y)\rightarrow (0,0)} \frac{e^{-x^{2}} - y}{x^{2}+y^{2}}$$ I have no idea about this question.Is the limit exist?

-

## 4 Answers

The numerator goes to $1$ as $(x,y) \to (0,0)$ and the denominator goes to $0$. It follows that the limit does not exist in $\mathbb{R}$. To show the quanity tends to $+\infty$ note that $$\frac{e^{-x^2}-y}{x^2+y^2} = \frac{1-x^2 + o(x^2+y^2)-y}{x^2+y^2} = \frac{1-x^2-y}{x^2+y^2} + o(1) \to +\infty$$ as $(x,y) \to (0,0)$ since the numerator is positive when $(x,y)$ is small, and the denominator is always nonnegative.

-
this doesn't mean that limit doesn't exist –  tinky_winky Mar 16 '13 at 15:16
@cf16 Yes it does. It might be clearer if you invert it: if $1/f(x,y) \to 0$ then $|f(x,y)| \to \infty$. –  Antonio Vargas Mar 16 '13 at 15:24
@AntonioVargas to be fair, the question could be asking whether the limit is $\pm \infty$ or whether there are paths which tend to each. I am updating my answer accordingly. –  nullUser Mar 16 '13 at 15:25
@cf16 Our disagreement comes from a difference in what we mean when we say the limit exists. For me, if the quantity tends to $\infty$ I do not say the limit exists because $\infty \notin \mathbb{R}$. If you are considering the function on $\overline{\mathbb{R}}$ with the order topology then you may say the limit is $\infty$ and that is fine too. –  nullUser Mar 16 '13 at 15:28
@cf16 Many people don't say it. –  1015 Mar 16 '13 at 15:52

$$\lim_{(x,y)\rightarrow(0,0)}\frac{e^{-x^2}-y}{x^2+y^2}=\lim_{x\rightarrow0}\bigg(\lim_{y\rightarrow0}\frac{e^{-x^2}-y}{x^2+y^2}\bigg)=\lim_{x\rightarrow0}\bigg(\frac{e^{-x^2}}{x^2}\bigg)=\infty$$

-
This could use more words. Really it only shows that the limit along a certain path is $\infty$. If you're using a set of definitions where "the limit is $\infty$" is subsumed in "the limit doesn't exist", this is sufficient to show that the limit doesn't exist, but if you're not it doesn't really prove anything... –  Micah Mar 16 '13 at 15:30
yes, you can't do second = this way I think –  tinky_winky Mar 16 '13 at 15:38
I just showed that using a certain path the limit tends to infinity. –  Occupy Gezi Mar 16 '13 at 15:47

I think the limit does not exist. Use Taylor series maybe more intuitive, the numerator is very closed to 1，and the denominator is closed to 0，is ∞.

-

just wanted to add picture: it seems that limit is $\infty$

-
want more pictures? different view? xD –  tinky_winky Mar 16 '13 at 15:41