# 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.