Finding the limit of a multivariable function

Question

So I'm trying to find this limit: $$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2y^2)}{(x^2+y^2)^{3/2}}$$

What I've tried so far is setting the value up for the sandwich theorem:

$$0\leq \Big|\frac{sin(x^2y^2)}{(x^2+y^2)^{3/2}}\Big|\leq\Big|\frac{sin(x^2y^2)}{(2x^2y^2)^{3/2}}\Big|$$ (using AM-GM inequality)

What troubles me is that I can't find a way to prove that this is less then/equal to zero.

Thanks in advance.

Answer 1

Generally not a good idea to introduce something in your denominator that will be unbounded in the required neighbourhood, if you have a choice. Why not change the numerator:

$$0\le\left|\frac{\sin(x^2y^2)}{(x^2+y^2)^{3/2}}\right|\le \frac{|x^2y^2|}{(x^2+y^2)^{3/2}}\le\frac{\left(\frac{x^2+y^2}{2}\right)^2}{(x^2+y^2)^{3/2}}=\frac{1}{4}\sqrt{x^2+y^2}\to 0$$

when $(x,y)\to(0,0)$.

Answer 2

I would rewrite it in polar coordinates:

$x = r \cos \theta$, $y = r \sin \theta$.

Then, the numerator would be $\sin(r^4 \cos^2\theta \sin^2\theta)$, which, in the neighborhood of $r=0$ would be smaller in magnitude than $\sin r^4$. Then, one can simply show that

$\textrm{lim}_{r \rightarrow 0} \frac{\sin r^4}{r^3}$

is zero.

Answer 3

Note that

$$\frac{\sin(x^2y^2)}{(x^2+y^2)^{3/2}}=\frac{\sin(x^2y^2)}{x^2y^2}\frac{x^2y^2}{(x^2+y^2)^{3/2}}\to 1\cdot 0=0$$

indeed $w=x^2y^2\to0$

$$\frac{\sin(x^2y^2)}{x^2y^2}=\frac{\sin w}{w}\to1\quad \quad\frac{x^2y^2}{(x^2+y^2)^{3/2}}\sim \rho \cdot f(\theta)\to 0$$