limit with two variable

Question

How to calculate:

$$ \lim_{(x,y) \to (0,0)} \frac{5x^6 + y^2}{x^3 + 2y} $$

I think the result should be $0$, but how do I prove it? I tried by the definition, but I could not resolve that.

I cannot use the different paths to prove that, because there are an infinity of paths and I would have to calculate them all, and that's impossible. I believe the only way is by definition.

@Mann, I get $\left({5k^5+k\over k^2+2}\right)t^3\to0$ as $t\to0$ from the path you suggest. — Barry Cipra, May 11 '15 at 19:45
But there are an infinity of paths,,, So only that path would not prove that limit is 0. I think the only way is using the definition of limit, but, like I said, I'm having problems there. — user239754, May 11 '15 at 20:08
Check the path $y=-x^3/2$. The quotient is not even defined along this path for any $(x,y)\ne(0,0)$. — Mark Viola, May 11 '15 at 20:26
@Dr.MV, it's standard (in some books, at least) to approach limits only along paths that stay in the domain of the function. — Barry Cipra, May 11 '15 at 20:38

Barry Cipra · Answer 1 · 2015-05-11T21:33:07.787

1

Hint: Let $y=-{1\over2}x^3+kx^p$, so that the denominator simplifies to $2kx^p$. Look for a $p$ that will cancel with what you get in the numerator.

Added later: If we approach $(0,0)$ along a path of the form $y=-{1\over2}x^3+kx^p$, we find that

$${5x^6+y^2\over x^3+2y}={5x^6+{1\over4}x^6-kx^{3+p}+k^2x^{2p}\over 2kx^p}={21\over8k}x^{6-p}-{1\over2}x^3+{k\over2}x^p$$

If we now let $p=6$, we get

$${5x^6+y^2\over x^3+2y}={21\over8k}-x^3+{k\over2}x^6\to{21\over8k}\quad\text{as }x\to0$$

Since different values of $k$ give different results, we conclude there is no limit.

(Remark: The proof that there is no limit could be written more tersely by setting $p=6$ from the get go. I presented things the way I did in part to show that the choice of $p$ is more or less forced.)

edited May 11 '15 at 21:33

answered May 11 '15 at 20:13

Barry Cipra

79,832

To the OP: your edit to your question indicates my hint didn't help. If so, please say so, and I'll give a more complete answer. – Barry Cipra May 11 '15 at 20:41
When you say cancel, you mean the numerator is now $0$, right? Well this, will give me $0$ for the value of the limit. And I cannot conclude anything from there, since the values I obtained using other paths are also $0$. I think the only way to solve this is using the definition of limit. – user239754 May 11 '15 at 21:08
1

Got it! Thanks for the help. – user239754 May 11 '15 at 21:50
@user239754, excellent. Glad to help. – Barry Cipra May 11 '15 at 21:58

score 0 · Answer 2 · answered May 11 '15 at 19:28

0

Don't assume that the limit is defined. See if you can find two paths y = y(x) such that the limits would not be equal.

answered May 11 '15 at 19:28

Sean Henderson

452

I know that and I've tried many of them, they all give me 0. But there are an infinity of ways, so I want to prove it. – user239754 May 11 '15 at 20:05

limit with two variable

2 Answers2