TITLE: 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)} $
QUESTION [5 upvotes]: 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

REPLY [3 votes]: 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$$\lim_{r\to0}r(\cos^3\theta+\sin^3\theta)=0$$