Calculate $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{\sin\left(x^{3}+y^{3}\right)}{\sin\left(x^{2}+y^{2}\right)} $ I have to calculate $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{\sin\left(x^{3}+y^{3}\right)}{\sin\left(x^{2}+y^{2}\right)} $
From wolfram calculator I know the limit is $ 0 $. The onl way I cant think of proving it is switch to polar, and to show that $ \lim_{r\to0}\frac{\sin\left(r^{3}\left(\cos^{3}\theta+\sin^{3}\theta\right)\right)}{\sin\left(r^{2}\right)} $ is $ 0$.
If I'll treat $ \theta $ as a constant and I'll get that the limit is zero, is that mean that from any direction that the function getting closer to zero, the limit is zero?
If so, I could show it using l'Hospital's rule and I guess it would be easy, but I'm not sure its legit.
Thanks in advance
 A: By your way the result is not immediately clear, as previously suggested by Mark Viola, we can use that
$$\frac{\sin\left(x^{3}+y^{3}\right)}{\sin\left(x^{2}+y^{2}\right)}=
\frac{\sin\left(x^{3}+y^{3}\right)}{x^3+y^3}
\frac{x^2+y^2}{\sin\left(x^{2}+y^{2}\right)}\frac{x^{3}+y^{3}}{x^{2}+y^{2}}$$
and using standard limits we reduce to evaluate the simpler
$$\lim_{\left(x,y\right)\to\left(0,0\right)}\frac{x^{3}+y^{3}}{x^{2}+y^{2}}$$

Note that assuming $\theta$ constant corresponds to take the limit by linear path (i.e. $y=mx$) and we can't conclude that the limit is zero in this way.
More in general we can't prove the limit existence by this way, as noticed by Mark Viola in the comments, "we can show a limit fails to exist by showing that the value of the limit has different results along different paths. But we don't show the existence by looking at the limit along paths".
We need to use some bounding and squeeze theorem to conclude, that is in this case
$$\frac{x^{3}+y^{3}}{x^{2}+y^{2}}=r (\cos^3 \theta + \sin^3 \theta) \to 0$$
since by squeeze theorem
$$\left|r (\cos^3 \theta + \sin^3 \theta)\right|=r\left|\cos^3 \theta + \sin^3 \theta\right| \le 2r \to 0$$
A: In your solution, you take $\theta$ constant with respect to $r$ and use l'Hospital. This proves that the limit is $0$ along straight line paths $y=x\tan\theta$ through the origin. But this doesn't prove that the limit exists in general. While here the limit exists and is equal to $0$, here is a question of mine posted about an year ago where I found that the limit may not exist despite having the same value along all straight line paths.

You can solve it like this: if $x^3+y^3=0$ i.e. along $y=-x$, the numerator is $0$ so the limit is $0$.
If $x^3+y^3\ne0$, then$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^3+y^3)}{x^3+y^3}\times\frac{x^2+y^2}{\sin(x^2+y^2)}\times\frac{x^3+y^3}{x^2+y^2}$$The first two terms are standard limits equal to $1$. You can write $\frac{x^3+y^3}{x^2+y^2}$ in polar coordinates as $r(\cos^3\theta+\sin^3\theta)$, so the limit is$$\lim_{r\to0}r(\cos^3\theta+\sin^3\theta)=0$$
A: You can just note that
$$
\lim_{(x,y)\to (0,0)}\dfrac{\sin(x^3+y^3)}{\sin(x^2+y^2)} = \lim_{(x,y)\to (0,0)}\frac{x^3+y^3}{x^2+y^2}
$$
and, regarding this last limit, since
$$
\left|\frac{x^3+y^3}{x^2+y^2} \right|\leq \frac{|x|^3+|y|^3}{x^2+y^2}\leq \frac{2(x^2+y^2)^{3/2}}{x^2+y^2}=\sqrt{x^2+y^2} \to 0
$$
we conclude that the original limit exists and is zero.
note: I'm using the fact that $|x|, |y| \leq \sqrt{x^2+y^2}$.
A: You can use inequality $\frac 12|u|\le |\sin(u)|\le |u|$ for small $|u|<1$ (it is true on a bit larger interval, but this is not very important).
$$0\le \dfrac{|\sin(x^3+y^3)|}{|\sin(x^2+y^2)|}\le 2\dfrac{|x^3+y^3|}{|x^2+y^2|}\le 2\dfrac{|x|^3+|y|^3}{x^2+y^2}\le 2\,\underbrace{\max(|x|,|y|)}_{\to 0}\ \underbrace{\dfrac{x^2+y^2}{x^2+y^2}}_1\to 0$$
