# I thought I proved the Limit here didn't exist…

So, this seems like it should be fairly simple, right? $$F(x,y) = {xy\over \sqrt{x^2+y^2}}$$

If one takes $\lim_{(x,y) \to (0,0)}F(x,y)$ along the line $y=x$, the limit simplifies to $\lim_ {x \to 0}\frac{x^2}{\sqrt{x^2+x^2}}$ which quickly reduces to $|x|/\sqrt{2}$, ie 0.

If one takes $\lim_{(x,y) \to (0,0)}F(x,y)$ along the parabola $y=x^2$, the equation simplifies to $\lim_{x \to 0} \frac{x^3}{\sqrt{x^2+x^4}}$ which evaluates to $0$ not as I thought $\infty$ by multiplication by $\frac{1 \over \sqrt{x^6}}{1 \over \sqrt{x^6}}$.

Therefore the limit could (as opposed to not, as I originally thought) exist. Alas, according to the book, this is incorrect - the limit is 0 (of course).

What is invalid about my approach and how should the problem be approached?

EDIT: I really appreciate all the answers, it has given me much to look into given time.... Dan's was most direct to the skillset this section is trying to develop.

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Your evaluation along the parabola is wrong. – DonAntonio Mar 14 '13 at 5:07
Sweet..... but I keep going over it, and I can't find anything wrong with my evaluation of the parabolic limit by multiplication .... anyone so kind as to enlighten me? :\ – user1833028 Mar 14 '13 at 5:28
Hint: $\sqrt{x^2+x^4} \sim \sqrt{x^2}$ – Jean-Claude Arbaut Mar 14 '13 at 6:19
Note that $\sqrt{2x^{2}}=|x|\sqrt{2}$ and not $x\sqrt{2}$. – T. Eskin Mar 14 '13 at 7:12
AAAAAAAAAhhhhhhhhh...... $1/\infty = 0$. Whoops. – user1833028 Mar 14 '13 at 7:26

$$\frac{x^3}{\sqrt{x^2+x^4}}=\pm\frac{x^2}{\sqrt{1+x^2}}\xrightarrow[x\to 0]{}0$$

In fact, check that

$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le\frac{|xy|}{|x|}=|y|\xrightarrow[(x,y)\to(0,0)]{}0$$

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You're on fire tonight, @Donantonio – Rustyn Mar 14 '13 at 5:07
I'm on insomnia tonight, Rustyn...*sigh*...But in a little while more I'm going to hit the sack! – DonAntonio Mar 14 '13 at 5:12
I know how that goes. – Rustyn Mar 14 '13 at 5:15

You can do it all at once using $|xy|\leq 2|xy|\leq x^2+y^2$ which is an often useful estimate following from $x^2+y^2-2|x||y|=(|x|-|y|)^2\geq 0$. Then $$0\leq |F(x,y)|\leq \frac{x^2+y^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}.$$ So $$\lim_{(x,y)\rightarrow(0,0)} F(x,y)=0$$ follows by squeezing.

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To find the limit, write $x=r\cos t$, $y=r\sin t$.

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if we take two path,like x goes to zero and y goes to zero,we see that limit exist,and both goes to zero,except where both x and y goes to zero,i think limit in this case exist – dato datuashvili Mar 14 '13 at 5:10
Hmm.... book does not mention any method like this... do you know of any link that I could use to see an example of this, sir? – user1833028 Mar 14 '13 at 5:12
If you make the substitution I suggested, the expression simplifies to $r\sin t\cos t$. This approaches $0$ as $(x,y)\to 0$, since $r\to 0$. – André Nicolas Mar 14 '13 at 5:12
listen to @AndréNicolas Nicolas – dato datuashvili Mar 14 '13 at 5:13
@user1833028: Sorry, I do not know an explicit reference. It is quite standard for the denominator $x^2+y^2$ and relatives. – André Nicolas Mar 14 '13 at 5:14
1. I would be more careful about your logic for the "parabola": The manipulations you've suggested are suited for a limit at infinity, not at zero.

2. The proof can be tricky, since you have to deal with two variables at once. My suggestion: Let $m = \max\{x,y\}$. Certainly $m \rightarrow 0$ as $x,y \rightarrow 0$. Can you find an expression, using $m$, which is greater than or equal to $f(x,y)$, and then show that this expression goes to zero?

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We should prove that: $$\forall\epsilon>0, \exists\delta>0,\forall (x,y): \left(0<||(x,y)-(0,0)||<\delta\longrightarrow\big|\frac{xy}{\sqrt{x^2+y^2}}-0|<\epsilon\right)$$ Firstly, see that $||(x,y)-(0,0)||<\delta$ is equvalant to saying that $\sqrt{x^2+y^2}<\delta$ and so both of $|x|$ and $|y|$ should be lass than $\delta$, so if we set $$z=\text{max}(|x|,|y|)$$ we get $$\big|\frac{xy}{\sqrt{x^2+y^2}}-0|=\frac{|x||y|}{\sqrt{x^2+y^2}}\leq\frac{zz}{\sqrt{z^2+0}}=z<\epsilon$$ Hence we need enough to take $\delta=\epsilon$.

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Very nicely done! +1 – amWhy Mar 15 '13 at 1:31