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$$f(x,y)=\frac{2^{xy}-1}{|x|+|y|}$$

I got to the conclusion that there is no limit, but I am not sure how to prove it.

AndreasT
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Alon
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1 Answers1

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Along the lines $\ x=0$, or $\ y=0$ the limit is zero; so the limit is zero or it doesn't exist.

Now, in the other cases, you can write the limit in this way: $$\ \lim_{(x,y)\to(0,0)}\frac{2^{xy}-1}{|x|+|y|}=\lim_{(x,y)\to(0,0)}\frac{2^{xy}-1}{xy}\cdot \frac{xy}{|x|+|y|}=$$ $$\ =\lim_{(x,y)\to(0,0)}\ln 2\cdot \frac{xy}{|x|+|y|}$$

Where I used the known limit $\ \lim_{t\to0}\frac{a^t-1}{t}=\ln a$, for $\ a>0,$ with $\ xy=t$

And now, it is easy to show that the last limit is zero:

$$\ |\frac{xy}{|x|+|y|}|≤\frac{|xy|}{|x|}=|y|\to 0$$

Mosk
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  • Can you please explain how do you get from $\frac{2^{xy}-1}{xy}$ to ln 2? – Alon Dec 06 '15 at 12:34
  • edited; and see also here, the last of 'notable special limits' https://en.wikipedia.org/wiki/List_of_limits – Mosk Dec 06 '15 at 13:34