Using polar coordinates to determine the limit

1. Jun 15, 2014

rmiller70015

Lim (x, y)->(0,0)(X^3+y^3)/(x^2+y^2)

The answer is -1, but I can't get it there. Here is what I did.
((Rcosx)^3 +(rsinx)^3)/((rcosx)^2+(rsinx)^2)
Then by factoring out a r squared from top and bottom I'm left with a denominator of (sin^2(x ) + cos^2 (x)) which simplifies to 1. And a numerator of:
Rcos^3(x)+rsin^3 (x)

This is about as far as I've gotten and I can't figure out how to get it to a -1.

2. Jun 15, 2014

Zondrina

The answer is certainly not -1. The answer is 0 as $r → 0$.

Conversely, if $\epsilon > 0$ is given, then choose $\delta = \frac{\epsilon}{2}$.

3. Jun 15, 2014

ehild

The limit is not -1. If it were, the function should get closer and closer to it, when (x,y) gets closer and closer to (0,0). If you choose x=y, f(x,x)=x →0. If you choose x=rcosθ, y=rsinθ, f=r(cos3θ+sin3θ) which goes to 0 if r→0, for any value of θ.

ehild