Prove $\lim_{(x, y) \to (0, 0)} f(x)$ exists if and only if $m + n > 2$

Question

Problem:

Let $m, n \in \mathbb{N}$. Show that $\lim_{(x, y) \to (0, 0)} \frac{x^{m}y^{m}}{x^{2} + y^{2}}$ exists if and only if $m + n > 2$.

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My try:

I'm really not too sure about how to prove this statement, but I think I can prove the forwards direction as follows:

I will show that if the limit exists, then we have $m + n > 2$; or equivalently, if $m + n \leq 2$, then the limit doesn't exist.

If $m + n \leq 2$ we must have $(m, n) = (1, 1)$. Then

$$\lim_{(x, y) \to (0, 0)} \frac{xy}{x^2 + y^2}.$$

Let $\{1/k, 1/k\}$ be a sequence converging to $(0, 0)$. Then

$$\lim_{(x, y) \to (0, 0)} = \frac{1}{k^2} \cdot \frac{k^2}{2} = 0.5,$$

but the sequence $\{1/k, 0\}$ also converges to $(0, 0)$ and

$$\lim_{(x, y) \to (0, 0)} 0 = 0, $$

and this gives two limit points, which cannot be true.

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I'm not sure how to prove the other direction. Can anyone please help me?

Answer 1

Your forward direction is fine, assuming that you don't think $0$ is a natural number. If you do, then a minor modification of your approach will take care of things.

For the converse direction, try working in polar coordinates. For example,

$$\frac{x^3 y^4}{x^2 + y^2} = r^5 \cos^3 \theta \sin^4 \theta.$$

The condition $m + n > 2$ should be readily apparent.

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Alternatively, use the AM-GM inequality. Note that

$$x^2 + y^2 = 2 \frac{x^2 + y^2}{2} \ge 2 \sqrt{x^2 y^2} = 2 |x| |y|$$

is strong enough to give

$$\left|\frac{x^m y^n}{x^2 + y^2}\right| \le \frac 1 2 |x|^{m - 1} |y|^{n - 1}.$$

Answer 2

Assume $m \geq n$. Let us show that $|x^{m}y^{n}| \leq x^{2}|y|$ whenever $|x|$ and $|y|$ are less than $1$. If $m \geq 2$ this is clear. If $m<2$ then $1\leq n\leq m<2$ which implies $m=n=1$ and $m+n=2$, a contradiction. Hence $|x^{m}y^{n}| \leq x^{2}|y|$. Now $|xy| \leq \frac 1 2(x^{2}+y^{2})$. It clear now that $f(x,y) \to 0$ as $(x,y) \to (0,0)$