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I want to find $\displaystyle\lim_{(x,y)\to(0,0)} \frac{xy}{x+y} = 0$. After trying different ways to approach $(0,0)$, I am fairly convinced the limit is $0$, but I need to prove it by definition, and I seem to be stuck. I want to prove that $\forall \epsilon >0, \exists \delta > 0 $ such that $0<\parallel(x,y)\parallel<\delta \implies |\frac{xy}{x+y}| < \epsilon$.

I'm having trouble with the denominator. I know that to get $|\frac{xy}{x+y}|$ to be less than something, I have to show that $|x+y|$ can be made greater than something, but I don't know what. Any suggestions?

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  • $\begingroup$ What if you (can you?) approach along the line $y = -x$? $\endgroup$ – Dylan Moreland May 4 '12 at 21:03
  • $\begingroup$ @Dylan: Well, the function is not even defined along that line. Does that matter? $\endgroup$ – Javier May 4 '12 at 21:05
  • $\begingroup$ @Javier: Well, it matters in-so-far as that means that your definition cannot be satisfied, since you can find points $(x,y)$ with $0\lt\lVert (x,y)\rVert \lt\delta$ but for which $\left|\frac{xy}{x+y}\right|\lt\epsilon$ does not hold... $\endgroup$ – Arturo Magidin May 4 '12 at 21:08
  • $\begingroup$ consider approaching on this curve $y=-x+x^3$, when $x$ is relatively small. $\endgroup$ – Yimin May 4 '12 at 21:10
  • $\begingroup$ @ArturoMagidin: Oh, I thought you could just sort of ignore it. This means that the limit doesn't exist, then? $\endgroup$ – Javier May 4 '12 at 21:13
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If you approach $(0,0)$ along the line $x=0$ the function has constant value $0$ and the limit is $0$.

But now suppose that you approach along a curve like $y=x^2-x$. Then $$\frac{xy}{x+y} = \frac{x^3-x^2}{x^2} =x - 1$$ and the limit as $x\to 0$ is...

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  • $\begingroup$ I guess this is a simpler way to know. How did you think of approaching along $y=x^2-x$? $\endgroup$ – Javier May 4 '12 at 21:17
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    $\begingroup$ @Javier: Experience, I'm afraid, not much more. $\endgroup$ – Arturo Magidin May 4 '12 at 21:23
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    $\begingroup$ @JavierBadia Clearly approach along a linear function won't do the job. So lets try the quadratic $y=ax^2 + bx$. Then we get that $\frac{ax^3 + bx^2}{ax^2 + (b+1)x} = \frac{ax^2 + bx}{ax +(b+1)}$. Note that if $b \neq -1$, then the limit will be zero. If $b=-1$, then we get $\frac{ax^2 - x}{ax} = \frac{ax-1}{a}$ and the limit is $-\frac1{a}$. So moving along the curve $y = ax^2 -1$ gives the limit $-\frac1a$ $\endgroup$ – user17762 May 4 '12 at 21:25
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The sum and product of two numbers are independent, you can assign them arbitrary values at will.

So while you make them both converge to zero, you can follow any $(s,p)$ curve, with arbitrary ratios.

For example, following an hyperbolic spiral $\rho=s/\theta$,

$$\frac{xy}{x+y}=\frac{s^2}{\theta^2}\frac{\theta}{s}\frac{\cos\theta\sin\theta}{\cos\theta+\sin\theta}=s\frac{\sin\theta}\theta\frac{\cos\theta}{\cos\theta+\sin\theta}.$$

This expression alternates between minus and plus infinity on every turn.

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Put $x=rcos\theta$ and $y=rsin\theta$ s.t. $$lim_{(r,\theta)\rightarrow (0,0)}\frac{rcos\theta.sin\theta}{cos\theta+sin\theta}=0$$ if and only if the denominator $cos\theta+sin\theta\ne 0$. However, approaching $r=0$ along $\theta=-\pi/4$ the given limit doesn't exist.

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    $\begingroup$ This is misleading, for reasons explained on the page in at least two posts, for almost 5 years when this was posted. $\endgroup$ – Did Mar 19 '18 at 12:34

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