Question

Does the limit lim(x,y)-->(0,0) (x^3 + y^3) / (x^2 + y^2) exist? If the limit exists, find its value.

Answer 1

So, this is a zero over zero limit. Do you know how to deal with these?

Answer 2

in two dimensions, yes.

Answer 3

It's much the same.

Answer 4

ok?

Answer 5

Not like I haven't tried it already...

Answer 6

Finding it in a text to make sure I don't tell you the wrong thing.

Answer 7

ok

Answer 8

I tried what I would have done it 2 dimensions but still came to 0/0 and then used l'hopital's to get =0 for every try, which would mean it exists... but I am almost certain that it does not.

Answer 9

Well, I can't find it in a text, but I did find this: http://www.physicsforums.com/showthread.php?t=112312

Answer 10

There's a little back and forth about what to do, but they are handling a very similar limit.

Answer 11

alright

Answer 12

Here is a thorough examination: http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/calculus/LimitsOfFunctionsOfTwoVariables.pdf

Answer 13

\[\left|\frac{x^3 + y^3}{x^2 + y^2}\right|=\left|\frac{(x+y)(x^2+y^2-xy)}{x^2+y^2}\right|=\left|\frac{(x+y)(x^2+y^2)+(x+y)(-xy)}{x^2+y^2}\right|\]

\[=\left|\frac{(x+y)(x^2+y^2)}{x^2+y^2}+\frac{(x+y)(-xy)}{x^2+y^2}\right|\]

\[=\left|x+y+\frac{(x+y)(-xy)}{x^2+y^2}\right|=\left|x+y-\frac{x^2y+y^2x}{x^2+y^2}\right|\]
\[\le |x+y|+\left|\frac{x^2y+y^2x}{x^2+y^2}\right|\le |x+y|+\left|\frac{x^2|y|+y^2|x|}{x^2+y^2}\right|\]
\[\le |x+y|+\left|\frac{x^2w+y^2w}{x^2+y^2}\right|\]

where \(w=\max\{|x|,|y|\}\)

\[=|x+y|+w\to 0\text{ as }(x,y)\to(0,0)\]