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

Question

Let me be honest, when some teacher ask me to calculate such limit, I always try to found two paths with different results to show that such limit don't exist.

But, I try four paths all fall at zero... and wolframalpha (internet boss (after math.stackexchange.com)) tells me this limit don't exit. Stewart's (book boss) calculus book tells me this limit is zero. But I don't know how to show this...

I hate limit calculus, because it's full of tricks... if you're not aware enough the day of your exam, you lost quickly some points...

Could you light my path?

thx to everybody!

A general useful inequality is $|x|,|y|\leqslant \sqrt{x^2+y^2}$, i.e. $|x|,|y|\leqslant \lVert (x,y)\rVert$. — Pedro, Oct 28 '13 at 01:58

score 11 · Answer 1 · edited Apr 13 '17 at 12:20

11

Related problems: I, II. Here is how you advance

$$ \Bigg| \frac{xy}{\sqrt{x^2+y^2}}-0 \Bigg |\leq \frac{|x||y|}{\sqrt{x^2+y^2}} \leq \frac{ \sqrt{x^2+y^2} \sqrt{x^2+y^2} }{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}< \epsilon =\delta .$$

Note:

$$ |x| \leq \sqrt{x^2+y^2},\quad |y| \leq \sqrt{x^2+y^2}. $$

edited Apr 13 '17 at 12:20

Community

1

answered Oct 28 '13 at 02:00

Mhenni Benghorbal

47,431

Sujaan Kunalan · Accepted Answer · 2013-10-28T01:55:20.733

9

Convert into polar coordinates: so $x=r\cos\theta\space$ and $y=r\sin\theta.$

$\displaystyle\lim_{(x,y)\to(0,0)} \dfrac{xy}{\sqrt{x^2+y^2}}\\=\displaystyle\lim_{r\to 0}\dfrac{r\cos\theta\cdot r\sin\theta}{\sqrt{r^2\cos\theta+r^2\sin\theta}}\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2(\sin^2\theta+\cos^2\theta)}} \text{Recall the Pythagorean Identity,}\sin^2\theta+\cos^2\theta=1.\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2}}\\=\displaystyle\lim_{r\to 0}\space r\sin\theta\cos\theta\\=0$

edited Oct 28 '13 at 01:55

answered Oct 28 '13 at 01:37

Sujaan Kunalan

10,944

1

thx, I didn't expected the full answer. But thank you for your "latex" effort – The Unholy Metal Machine Oct 28 '13 at 02:06
1

You're welcome. :) – Sujaan Kunalan Oct 28 '13 at 02:07
double thx, because that limit was in my exam this morning – The Unholy Metal Machine Oct 28 '13 at 16:37
2

This doesn't show that the function is continuous at (0, 0). There is always a possibility that the limit as (x,y) goes to (0, 0) along a curve (a path that is not a straight line) is not 0. So this is not a complete proof. – Clyde Sep 06 '15 at 23:14
@Nick How would you fix the proof? It looks like they fixed $\theta $ and let $r$ goes to zero. – john Oct 10 '17 at 01:33

score 5 · Answer 3 · answered Oct 28 '13 at 01:49

5

You could do the following: $x^2+y^2\geq 2|xy|$, so $\sqrt{x^2+y^2}\geq \sqrt{2|xy|}$, therefore $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq\frac{|xy|}{\sqrt{2|xy|}}=\frac{\sqrt{|xy|}}{\sqrt{2}},$$ which converges to $0$ as $x,y\to 0$.

answered Oct 28 '13 at 01:49

detnvvp

8,237

score 2 · Answer 4 · answered Oct 28 '13 at 01:32

2

The hint is to use the standard trick of the polar coordinates thing:

Put $x = r \cos \theta $, $y = r \sin \theta $.

Alternatively, you can see what happens if you approach the limit through the the line $y = x $.

answered Oct 28 '13 at 01:32

ILoveMath

10,694

score 0 · Answer 5 · answered Oct 28 '13 at 01:37

0

There really isn't any "trick" here; just applying standard absolute value estimates and polar coordinates. As @Citizen mentioned, polar coordinates turns the inside of your limit to

$$\dfrac{r^2\cos\theta\sin\theta}{r}.$$

Try it from there...

answered Oct 28 '13 at 01:37

Patch

4,245

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

5 Answers5

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