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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?

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

2 Answers 2

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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).

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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 $\theta$.

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  • $\begingroup$ what if $\sin(2\theta)= 1/\rho^2$? $\endgroup$
    – michek
    Mar 22, 2015 at 17:21
  • $\begingroup$ This only considers very specific sublimits. See this. This isn't right. $\endgroup$
    – Git Gud
    Mar 22, 2015 at 19:00

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