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Can anyone give me some hints on how to solve this limit: $\lim_{(x,y)\to (0,0)} \frac{e^x-e^y}{x^2y^2}$ I have considered using power series of e or changing to polor coordinates but neither have yielded any results that can prove whether this limit exists or not. Any hints or suggestions would be helpful. Thank you.

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    $\begingroup$ A usual tactic to tackle these limits is to check the limit along lines of the form $y=mx$. $\endgroup$ – player3236 Oct 21 at 15:16
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We have that for $x=y=t$

$$\frac{e^x-e^y}{x^2y^2}=\frac{e^t-e^t}{t^4}=0$$

and for $x=t$, $y=-t$ with $t \to 0^+$

$$\frac{e^x-e^y}{x^2y^2}=\frac{e^t-e^{-t}}{t^4}=\frac{2}{t^3e^t}\frac{e^{2t}-1}{2t} \to \infty$$

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  • $\begingroup$ Sorry, my mistake. $\endgroup$ – Andrei Oct 21 at 15:25
  • $\begingroup$ @strawberry-sunshine I'm assuming $t\to 0^+$ and since in this case the expression tends to $\infty$ this suffices to conlcude that the limit for the given function doesn't exist. Of course we can only use the second case to conclude considering the limit for $1/t^3$ as $t\to 0$. $\endgroup$ – user Oct 21 at 15:29
  • $\begingroup$ Yes, I agree. Sorry, I missed $t \to 0^+$ earlier. $\endgroup$ – strawberry-sunshine Oct 21 at 15:32
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    $\begingroup$ @strawberry-sunshine That's fine, your idea is indeed very good and smart! $\endgroup$ – user Oct 21 at 15:50
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Consider the line $y = -x$, along which the limit becomes $$\lim_{x \to 0} \frac{e^x - e^{-x}}{x^4}$$ which clearly does not exist. $$\lim_{x \to 0^+} \frac{e^x - e^{-x}}{x^4} = \infty \text{ and }\lim_{x \to 0^-} \frac{e^x - e^{-x}}{x^4} = -\infty $$You may try to show this using the Maclaurin expansion of $e^x$, or by some other way.

If the original limit did exist, it would exist for $y = -x$ too.

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