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$$\lim_{(x,y)\to 0,0}\frac{e^{xy}-1}y$$

I'm new to this. I do not think you can use epsilon-delta def. I'm also confident that the limit approaches $0$. I just don't know how to prove it.

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There was a silly comment by me here before. If you saw it, try to forget. –  André Nicolas Mar 7 '13 at 3:05
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2 Answers

up vote 2 down vote accepted

Hints:

Let $z = xy$ and see if you can make progress. You should arrive at a limit of zero.

You can also do this using the epsilon-delta definition.

Update

Compute:

$$\displaystyle \tag 1 \lim_{(x,y)\to 0,0}\frac{e^{xy}-1}{y}$$

Let $z = xy$ and rewrite $(1)$, as:

$$\tag 2 \frac{e^{xy}-1}{y} = x \cdot \frac{e^{z} -1}{z}$$

Now, $\lim_{(x,y)\to 0,0} x = 0$, futhermore, we have (use L'Hôpital's rule):

$$\lim_{z\to 0}\frac{e^{z} -1}{z} = 1$$

By the product rule of limits, we have:

$$\lim_{(x,y)\to 0,0}\frac{e^{xy}-1}{y} = \left(\lim_{(x,y)\to 0,0} x \right) \cdot \left(\lim_{z\to 0}\frac{e^{z}-1}{z} \right) = (0) \cdot (1) = 0.$$

I'll let you tackle the epsilon-delta variant as it is not bad.

Regards.

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Can't do either. It's not clear how to use epsilon-delta when dealing with e (or other transcendentals), probably because of my lack of experience. I have no idea why'd you sub xy for z. It looks eerily like the def of derivative, but that's all I got. Thanks for the quick reply btw. –  orbis Mar 7 '13 at 3:36
    
@orbis: see update. Regards –  Amzoti Mar 7 '13 at 4:12
    
Thank you. You definitely helped a lot. I did figure out a delta-epsilon proof although I am not sure if it is sufficiently strong but it will have to do. –  orbis Mar 7 '13 at 6:02
    
@orbis: You can post that as a new question with your solution and ask for feedback. –  Amzoti Mar 7 '13 at 6:10
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Nice follow-up and updates as needed. I really appreciate users who "converse" with OPs, (when the opportunity to do so presents itself!) –  amWhy Apr 24 '13 at 0:44
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Easier, without L'Hospital's rule. As $(x,y) \to 0, \ e^{xy}=1+xy+O((xy)^2)$, hence your expression becomes $$ \lim_{x,y \to 0}\frac{e^{xy}-1}{y}=\lim_{x,y \to 0} \frac{xy+O((xy)^2)}{y}=\lim_{x,y \to 0}x+O(x^2y)=0 $$

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