Xrfgtjtk

I struggle with this problem for hrs.

Point out whether the limit of the function exists or not.

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

I tried to $$lim_{x,yto 0,0} frac{sin(x^3)x}{(x^3)}=1(0)=0$$

but I can't find any anther path to find the different limit of this question.

We have $sin zleq |z|$ for every $z$ close enough to zero.
It follows that
$$left |frac{sin(x^3+y^3)}{x^2+y^2}right|leq frac{|x^3+y^3|}{x^2+y^2}leq frac{|x|^3}{x^2}+frac{|y|^3}{y^2}=|x|+|y|$$

But since the limit of $|x|+|y|$ goes to zero as $(x,y)to (0,0)$ then so is $frac{sin(x^3+y^3)}{x^2+y^2}$.

Well, in limits in which your are going to zero ($hto 0$) it is very useful the inequality $|sin(h)|leq h$. In fact:
begin{align} limlimits_{(x,y)to(0,0)} left|frac{sin(x^3+y^3)}{x^2+y^2}right| & leq limlimits_{(x,y)to(0,0)} left|frac{x^3+y^3}{x^2+y^2} right|\ text{use now, polar coordinates: $x=rhocos(theta)$, $y=rhosin(theta)$}\ leq limlimits_{rhoto 0} left| rho frac{cos(theta)^3+sin(theta)^3)}{cos(theta)^2+sin(theta)^2}right|to 0end{align}
since $cos(theta)^2+sin(theta)^2=1$ and $cos(theta)^3+sin(theta)^3)$ is bounded.

We have that

$$frac{sin(x^3+y^3)}{x^2+y^2}=frac{sin(x^3+y^3)}{x^3+y^3}cdotfrac{x^3+y^3}{x^2+y^2}to 0$$

indeed by $t=x^3+y^3to 0$ we have

$$frac{sin(x^3+y^3)}{x^3+y^3}= frac{sin t}{t}to 1$$

and for example by polar coordinates

$$frac{x^3+y^3}{x^2+y^2}=r(cos^3theta+sin^3theta)to 0$$