Finding $ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $

Question

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

If I pick $ x = 0$ I get:

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

So if the limit exists it must be $0$

Now for ${(x,y) \to (0,0)}$ I have $xy \to 0$

So I can use the Taylor series of $sin(t) = t + o(t)$ where $t \to 0$

$$0 \leq \frac{\sin^2(xy)}{3x^2+2y^2} = \frac{x^2y^2 + 2o(x^2y^2) + o (x^2y^2)}{3x^2 + 2y^2} =$$

$$\frac{x^2y^2 + o (x^2y^2)}{3x^2 + 2y^2} = $$

Now I can use the polar coordinates:

$$\frac{\rho^4 \cos^2(\theta)\sin^2(\theta) + o (\rho^4 \cos^2(\theta)\sin^2(\theta))}{3\rho^2 \cos^2(\theta) + 2\rho^2 \sin^2(\theta)}=$$

$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 \cos^2(\theta) + 2 \sin^2(\theta)}=$$

$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 - 3 \sin^2(\theta) + 2 \sin^2(\theta)}=$$

$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 - \sin^2(\theta)}\leq$$

$\frac{\rho^2}{2} \to 0$ for $\rho \to 0$

I would like to know if it is solved in the right way

Answer 1

Or $\sin^2(xy)\leq(xy)^2$ and $3x^2\geq 2x^2$ thus $\frac{\sin^2(xy)}{3x^2+2y^2}\leq \frac {(xy)^2}{2(x^2+y^2)}=xy\cdot \frac {xy}{2(x^2+y^2)}\leq xy\cdot \frac {1}{4}\to 0$ because $\frac {xy}{x^2+y^2}\leq \frac {1}{2}\Leftrightarrow 2xy\leq x^2+y^2\Leftrightarrow 0\leq (x-y)^2$.

Answer 2

We have $\;3x^2+2y^2\geq 2\sqrt{6}|xy|\;$ by AM-GM, thus we have

$\dfrac{\sin^2(xy)}{2\sqrt{6}|xy|}=\left(\dfrac{\sin(xy)}{xy}\right)^2\dfrac{|xy|}{2\sqrt{6}}\to0\;$ for $\;(x,y)\to(0,0)\;.$

I think your proof is ok. You can read also my proof if you want, it is a little bit shorter.