# Regarding $\lim \limits_{(x,y) \to (0,0)} xy\exp\left(\frac{xy}{x^2+y^2}\right)$

I am trying to prove that the limit
$$\lim_{(x,y) \to (0,0)} xy\exp\left(\frac{xy}{x^2+y^2}\right)$$

does not exist, by considering the hyperbolae $\gamma^+ =\{y= 1/x\}$,and $\gamma^-= \{y=-1/x\}$. However I get an expression of the form $$\pm \exp(\frac{1}{ x^2+1/x^2})$$ which goes to $\pm 1$ but is not independent of $x$. Is there a better way to do this?

• Maybe I'm being dumb, but what use is considering these hyperbolas if you're trying to approach (0,0)? Mar 22, 2015 at 17:11

When considering different sublimits, you need to take paths that approach the desired point. The paths $t\mapsto \left(t,\dfrac1 t\right)$ do not approach the origin $(0,0)$ anywhere.
The limit actually exists and this can be proven by showing that $(x,y)\mapsto \exp\left(\frac{xy}{x^2+y^2}\right)$ defined for $(x,y)\neq (0,0)$ is bounded.
In fact, given $(x,y)$, it holds that $-\left(x^2+y^2\right)\leq 2xy\leq x^2+y^2$.
It's not hard to transform this inequalities to prove that the given function is bounded. (Remember than $\exp$ is increasing and division by positive numbers preserves inequalities).
Are you sure? If $x=\rho\cos\theta$ and $y=\rho \sin \theta$ we have $$\lim_{(x,y) \to (0,0)} xy\exp\left(\frac{xy}{x^2+y^2}\right)=\lim_{\rho\rightarrow 0}\frac{\rho^2}{2} \sin2\theta \exp \left(\frac{\sin2\theta}{2}\right)$$ this limit is equal to 0 for all value of $\theta$.
• what if $\sin(2\theta)= 1/\rho^2$? Mar 22, 2015 at 17:21