Proving that $\lim\limits_{(x, y)\to (0, 0)} \frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$ doesn't exist

Question

I want to prove that $\displaystyle \lim_{(x, y)\to (0, 0)} \frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$ doesn't exist

I denoted $f(x, y):=\frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$. I tried looking at $f(\frac{1}{n}, \frac{1}{n})$ and other similar sequences, but no result. No matter what sequence I chose I got that the limit is $0$. However, WA says that this doesn't exist. What sequence should I pick?

Answer 1

As Maxim pointed out the limit actually exists and is equal to 0. I give an alternative proof for that now. It suffices to show, that for |x|,|y| sufficiently small there holds

$$\left| \frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\right| \le \sqrt{2}\max(\sqrt{|y|},|x|^{\frac{1}{3}}).$$

This we see at follows. First by Cauchy-Schwart and the triangle inequality there holds

$$\frac{1}{\sqrt{2}}(|x|+|y|)\le \sqrt{x^2+y^2}\le |x|+|y|.$$

We assume in the following w.o.l.g. that $1>>x,y>0$. Then,

$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^3}{(x^4+y^6)(x+y)}.$$

First assume, that $x\ge y$. Then

$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^3}{(x^4+y^6)(x+y)}\le \sqrt{2}\frac{x^2y^3}{x^4} =\le \sqrt{2} \frac{y^3}{x^2}\le \sqrt{2} x$$ and the above statement is fulfilled. We now assume in the following, that $x\le y$. Then

$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^2}{x^4+y^6}.$$

Now we consider the two cases $x^4\le y^6$ and $x^4\ge y^6$. In the first case,

$$\frac{x^3y^2}{x^4+y^6} \le \frac{x^3y^2}{y^6}\le \frac{x^3}{y^4} \le \frac{y^\frac{3*6}{4}}{y^4} = \sqrt{y}.$$

In the remaining case,

$$\frac{x^3y^2}{x^4+y^6} \le \frac{x^3y^2}{x^4} = \frac{y^2}{x}\le x^\frac{1}{3}.$$

Therefore, the statement is proved.

The following reasoning does not yield the statement, as Maxim pointed out:

Set $y=y(x)=x^q$ with $q>0$. You can try to tune $q$, such that the numerator dominates the denominator. By the above, $q$ has to be smaller than 1. For the denominator there holds

$$(x^4+x^{6q}) \sqrt{x^2+x^{2q}} = x^{\min(4,6q)}(1+O(1)) x^{min(1,q)} (1+O(1))$$

for $x\to0$. E.g., for $q=4/6$ we have

$$x^{3+3q} = x^5 > x^{28/6}(1+o(1)) = x^{7q}(1+o(1)).$$

Thus for that choice, the sequence will diverge (This last deduction is wrong).

Answer 2

In this type of exercises of "non-existence of limits" you might prove it by 'Reductio ad absurdum':

1. You suppose that the limit exists. Now you are going to calculate its value.
2. In order to calculate it, you aproach the point $(0,0)$ in the $\mathbb{R}^2$ plane by one way (by $x=y,\ x=y^2,\ x=\frac{1}{n}\ y=\frac{1}{m}$, whatever) and it comes that its value is anything, igamine $1$.
3. Now, you take another aproach, different than the previous onne. The limit also exists, but its value its something different, imagine $0$.
4. So, the limit 'exists' and have two values. By the uniqness of the limit in $\mathbb{R}^n$, this can't happen, so it is a contradiction. So, your initial asumption "The limit exists" is false.

Conclude that the limit doesn't exists.