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 $\lim_{t\to 0}\sin t / t=1$.
$$\lim_{(x,y)\to (0,0)} \frac{3x^2\sin^2y}{2x^4+2\sin y^4}$$
 A: Consider various paths that approach $(0,0)$. For instance,
$$\lim_{(x,y)\to (0,0)}\frac{3x^2\sin^2y}{2x^4+2\sin(y^4)}$$ along the line $y=0$ becomes
$$\lim_{x\to 0}\frac{3x^2\cdot 0}{2x^4}=0$$
However, along the line $y=x$ the limit is 
$$\lim_{x\to 0}\frac{3x^2\sin^2x}{2x^4+2\sin(x^4)}=1$$ I evaluated this limit numerically. This shows that the limit does not exist. You can evaluate it by l'Hopital, careful algebra and the hint.
A: The first approximation
I would use is
$\sin(x)
\approx x$
for small $x$.
Therefore
$\frac{3x^2\sin^2y}{2x^4+2\sin y^4}
\approx \frac{3x^2y^2}{2x^4+2y^4}
= \frac{3(x/y)^2}{2(x/y)^4+2}
$
(dividing by $y^4$).
Therefore,
if $x/y=c$,
this approaches
$\frac{3c^2}{2c^4+2}
$,
which can take on a variety of values.
A: $$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{x^2sin^2(y)}{x^4+siny^4})$$ 
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
