Finding $\lim_{(x,y)\to (0,0)} \frac{3x^2\sin^2y}{2x^4+2\sin y^4}$ I had following limit of two variables as a problem on my calculus test. How does one show whether the limit below exists or does not exist? I think it does not exist but I was not able to show that rigorously. There was a hint reminding that $\lim_{t\to 0}\sin t / t=1$.
$$\lim_{(x,y)\to (0,0)} \frac{3x^2\sin^2y}{2x^4+2\sin y^4}$$

$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{x^2sin^2(y)}{x^4+siny^4})$$ 
If Dividing by  $y^4$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{\frac{x^2sin(y)sin(y)}{y^2yy}}{\frac{x^4}{y^4}+\frac {sin(y^4)}{y^4}})$$
Know:$\lim_{t\to0}\frac{sint}{t}=1$
$(y^4=t)$,$(y=t)$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{\frac{x^2}{y^2}}{\frac{x^4}{y^4}+1})$$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{x^2y^2}{x^4+y^4})$$
Way $y=x$
$$\lim_{(x,y)\to (0,0)}\frac{3x^4}{4x^4}=\frac{3}{4}$$
Way $Y=0$ limit hint to 0
Not exist
