# Multivariable Calculus proof!

1. Sep 21, 2011

### _Steve_

I need to show that

limit (|x|^a*|y|^b) / (|x|^c+|y|^d) = 0
(x,y)->(0,0)

when a,b>=0; c,d>0; with a/c + b/d > 1

Does anyone have some tips for starting off the proof?

2. Sep 21, 2011

### _Steve_

Show that if $a, b \geq 0$ and $c, d > 0,$ with $\frac{a}{c} + \frac{b}{d} > 1,$ then:

$lim_{\vec{x}\rightarrow\vec{0}} \frac{|x|^{a}|y|^{b}}{|x|^{c}+|y|^{d}} = 0$

3. Sep 21, 2011

### HallsofIvy

Staff Emeritus
Do you mean $\lim_{x\to 0}$ or $\lim_{(x,y)\to 0}$. If the former, it is pretty easy! As x goes to 0, for any y, the numerator goes to 0. If y is not 0, the denominator goes to $|y|^d$ and, of course, $0/|y|^d= 0$. If y is 0, then the function is 0 for all x so its limit is still 0.

If the problem is $\lim_{(x,y)\to 0}$,l that's a bit harder.

4. Sep 21, 2011

### _Steve_

It's the second one. I tried to put the vector lines over the x and 0 lol.

Yeah I'm not sure how to start this! Should I try using Squeeze theorem with something? Or the definition of a limit?

5. Sep 22, 2011

### _Steve_

I figured it out, decided to post the answer just in case someone else has the same kind of question sometime...
basically just change the numerator to:
(|x|^c)^(a/c) (|y|^d)^(b/d)
and use the inequalities:
|x|^c <= |x|^c + |y|^d
|y|^d <= |x|^c + |y|^d
then cancel out and use squeeze theorem