Homework Help: Multivariable limit (x,y) to (0,0)

1. Jan 24, 2012

kottur

1. The problem statement, all variables and given/known data

Calculate the limit:

$lim_{(x,y)\rightarrow(0,0)}=\frac{4\sqrt{x^{14}+y^{14}}}{(x^{2}+y^{2})^{3}}$

3. The attempt at a solution

I think I'm supposed to use algebra for this but I can't seem to figure anything useful out.

I've found this:

$x^{14}+y^{14}=(x^{2}+y^{2}) (x^{12}-x^{10} y^{2}+x^{8} y^{4}-x^{6} y^{6}+x^{4} y^{8}-x^{2} y^{10}+y^{12})$

but I'm not sure that helps me at all.

I know the limit exists and it's zero because I've plotted it.

Any help would be greatly appreciated.

2. Jan 24, 2012

SammyS

Staff Emeritus
Change this to polar coordinates.

3. Jan 24, 2012

kottur

I plugged in $x=rcos(\theta)$ and $y=rsin(\theta)$ and after a little bit of cleaning up I got:

$lim_{(r,\theta)\rightarrow(0,0)}=4r\sqrt{cos^{14}(\theta)+cos^{14}(\theta)}$

But the limit is supposed to be one... I'm note sure I'm doing this right, or maybe I just don't understand this as well as I wanted.

4. Jan 24, 2012

SammyS

Staff Emeritus
That should be $\displaystyle\lim_{r\to 0}\ 4r^5\sqrt{\cos^{14}(\theta)+\sin^{14}(\theta)}\,.$

θ does not go to zero.

The limit is zero, not 1.

5. Jan 24, 2012

kottur

Yep of course the limit is zero, sorry about that.

This is how I calculated my result:

$lim_{(x,y)\rightarrow(0,0)}\frac{4\sqrt{x^{14}+y^{14}}}{(x^{2}+y^{2})^{3}}=lim_{r\rightarrow0}\frac{4\sqrt{r^{14}cos^{14}(\theta)+r^{14}sin^{14} (\theta )}}{(r^{2}cos^{2}(\theta)+r^{2}sin^{2})^{3}(\theta)}=lim_{r\rightarrow0}\frac{4\sqrt{r^{14}}\sqrt{cos^{14}(\theta)+sin^{14}(\theta)}{}}{(r^{2}(cos^{2}(\theta)+sin^{2}(\theta))^{3}}=lim_{r\rightarrow0}\frac{4r^{7}\sqrt{cos^{14}(\theta)+sin^{14}(\theta)}}{r^{6}}=lim_{r\rightarrow0}4r\sqrt{cos^{14} (\theta)+sin^{14}(\theta)}$

I don't see what I did wrong here.

Also why don't I put $\theta\rightarrow0$? I understand that $\theta$ goes in a period but shouldn't it approach anything? Maybe $2\pi$?

6. Jan 24, 2012

Dick

As along as r->0 the limit is still 0. The trig term bounded by $\sqrt 2$ for any value of $\theta$

7. Jan 24, 2012

kottur

What do you mean by that?

Thank you for all the help.

8. Jan 24, 2012

Dick

I mean that $\sqrt(\sin^{14} \theta+\cos^{14} \theta)$ is bounded as theta goes from 0 to 2*pi. It never gets larger than sqrt(2). If it went to infinity for some value of theta that could be a problem for the limit. The value of your function is between 0 and 4*r*sqrt(2). So as r->0 it must go to zero.

9. Jan 24, 2012

kottur

I get it now. Thank you.