Limit of two variables, proposed solution check: $\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}$

Question

Does this solution make sense,

The limit in question:

$$ \lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}} $$

My solution is this:

Suppose, $$ \sqrt{x^2+y^2} < \delta $$ therefore $$xy<\delta^2$$ So by the Squeeze Theorem the limit exists since $$\frac{xy}{\sqrt{x^2+y^2}}<\frac{\delta^2}{\delta}=\delta$$

Is this sufficient?

If you want to use that approach to show that the limit is $0$, then you'll need to know that $|xy|<\delta^2,$ not that $xy<\delta^2$. — Cameron Buie, Jun 11 '13 at 00:00
Be careful. If $\sqrt{x^2+y^2}<\delta$, then it's certainly not the case that $\dfrac1{\sqrt{x^2+y^2}}<\dfrac1{\delta}$. You need to be careful with your inequalities!! — Ted Shifrin, Jun 11 '13 at 00:05
Let $y=kx$ for constant k that defines the direction of the limit. Then the limit becomes $$\frac{k}{\sqrt{1+k^2}}\lim_{x\rightarrow 0}\frac{x^2}{x}=0 $$ Therefore the limit exists unless $k=\pm i$ — Abhimanyu Pallavi Sudhir, Jun 11 '13 at 04:23
If you are using $\epsilon-\delta$ proof, then this technique, makes the problem falls apart. — Mhenni Benghorbal, Jun 11 '13 at 04:32

score 4 · Answer 1 · answered Jun 11 '13 at 00:03

4

Consider the continuous curve $C_{\alpha}(t)=(t,\alpha t)\subset\mathbb R^2$. Obviously $\lim C_{\alpha}(t)=(0,0)$ when $t\to 0$ and $$\lim_{t\to 0} f(C_{\alpha}(t))=0$$ so the limit of $f$ is probably exists. Now note that if $$||(x,y)-(0,0)||<\delta$$ then we have equivalently $$\sqrt{x^2+y^2}<\delta$$ and so $$|x|<\delta,~~|y|<\delta$$ Therefore if $z=\text{max}\{|x|,|y|\}$ so $z<\delta$ and $$\big|\frac{xy}{\sqrt{x^2+y^2}}-0\big|=\frac{|x||y|}{\sqrt{x^2+y^2}}\le \frac{z^2}{\sqrt{z^2+0}}=z<\delta$$

answered Jun 11 '13 at 00:03

Mikasa

67,374

$+ 1 \land \ddot\smile$ – amWhy Jun 11 '13 at 00:06
1

if I use $\dfrac{|xy|}{\sqrt{x^2+y^2}}= \le \dfrac{x^2+y^2}{2\sqrt{x^2+y^2}}=\dfrac{\sqrt{x^2+y^2}}{2}$,is it OK also? – chenbai Jun 11 '13 at 03:22
@chenbai: seems right. but mine is better ;-) – Mikasa Jun 11 '13 at 05:11
thanks. I learn your method. – chenbai Jun 11 '13 at 05:18

score 2 · Answer 2 · answered Jun 11 '13 at 04:06

2

Here's a more direct solution.

We know $|x|,|y|\le\sqrt{x^2+y^2}$, so if $\sqrt{x^2+y^2}<\delta$, then $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le\frac{\big(\sqrt{x^2+y^2}\big)^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}<\delta.$$

answered Jun 11 '13 at 04:06

Ted Shifrin

115,160

score 1 · Answer 3 · answered Jun 10 '13 at 23:59

1

I'm not sure I really follow your solution well - it doesn't seem to make sense to me (edit: see Cameron Buie's comment). Why not use polar coordinates by making the substitution $x = r\cos \theta, y = r \sin \theta$ though?

answered Jun 10 '13 at 23:59

Alex Wertheim

20,278

You can use polar coordinates, provided you give an estimate that is universal independent of $\theta$. See my detailed response at http://math.stackexchange.com/questions/405884/lhopitals-rule-and-multivariable-limits/405947#405947. – Ted Shifrin Jun 11 '13 at 00:12
+1 @AWertheim : I would definitly use polar coordinates: $f(r\cos\theta,r\sin\theta)=r\cos\theta\sin\theta\rightarrow 0$ when $r\rightarrow 0$ as $\cos\theta\sin\theta$ is bounded. – Avitus Jun 11 '13 at 09:24

score 0 · Answer 4 · answered Jun 11 '13 at 05:04

0

Another direct approach is to note that the square of the reciprocal of your function is $\frac{1}{x^2}+\frac{1}{y^2}$.

answered Jun 11 '13 at 05:04

Bill Kleinhans

1,458

Limit of two variables, proposed solution check: $\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}$

4 Answers4