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I'm having some issues with $\varepsilon$-$\delta$ proofs of limits with more than one variable. I understand the $\varepsilon$-$\delta$ definition of a limit, but I don't know how to deal with multiple variables.

Here's a simple example: prove the following limit, if it exists, using the epsilon-delta definition: $$ \lim_{(x,y)\to (1,2)}\frac{x^2}{x+y} = a \iff \forall(\varepsilon>0)\, \exists(\delta) \left[ |(x,y)-(1,2)|<\delta \implies \left|\frac{x^2}{x+y}-a\right|<\varepsilon \right] $$

Obviously this limit is $1/3$, but how do I prove it?

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Let $x= 1 +t\;\;y=2+s$ Let $s^2 +t^2<\delta^2$ $$\dfrac{x^2}{x+y}=\dfrac{1}{3+t+s} +\dfrac{2t+t^2}{3+t+s}$$ $$|\dfrac{2t+t^2}{3+t+s}| \leq |\dfrac{2t+t^2}{3-2\delta}|\leq3\dfrac{\delta}{3-2\delta}\leq3\delta$$ If $\delta<1.$

$$|\dfrac{1}{3+t+s}-\frac13|=|\dfrac{t+s}{3(3+t+s)}|\leq\dfrac{2\delta}{3(3-2\delta)}\leq\dfrac{2\delta}{3}$$ Using this you can complete the proof.

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  • $\begingroup$ Set $\delta=\frac{3\epsilon}{22}$, so $|\frac{x^2}{x+y}-\frac{1}{3}|=|\frac{t^2+2t+1}{t+s+3}-\frac{1}{3}|=|\frac{t^2+2t}{t+s+3}+\frac{1}{t+s+3}-\frac{1}{3}|\leq |\frac{t^2+2t}{t+s+3}|+|\frac{t+s}{3(3+t+s)}|\leq 3\delta +\frac{2\delta}{3}=\frac{11\delta}{3}=\frac{\epsilon}{2}<\epsilon$. $\endgroup$ – user54448 Mar 4 '13 at 18:56
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Actually, It's quite obvious. Whenever we have multiple variables involved, look for the interval that the variables are in, and we'll able to find a bound (upper or lower) for the variables.

For example, in your example, the interval for (x,y) is (1,2). Thus, I claim x < 1 and y < 2 respectively, and note the inequality are strict, since this interval is not closed. So then, when I plug these bounds into the last part of your implication, I get 1 / (1+2) = 1/3 < $\epsilon$

I hope this helps.

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  • $\begingroup$ I'm not sure I follow, I would really appreciate it if you could clear up some things: What exactly do you mean when you say the interval for (x,y) is (1,2)? And what is your justification for omitting the cases when x>1 or y>2? Additionally, why are you allowed to plug in the bounds into the last part of the implication (since we want not the actual value at that point, but what it is tending to as we approach that point)? $\endgroup$ – user54448 Mar 4 '13 at 21:18
  • $\begingroup$ @divergent : Actually, I'm a bit shaky in notation. I should not say the "interval" for (x,y), but rather the "controller" for (x,y). In other words, x approaches 1 and y approaches 2. If x>1 or y>2, we just need to adjust the inequalities around, since they are still "controlled" by (1,2). So assuming x<1 and y<2 is enough to finish the proof $\endgroup$ – Cecile Mar 5 '13 at 22:00

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