<s>[INST] $\lim_{(x,y)\rightarrow (0,0)}\frac{x^5-\sin(y^3)}{4x^4+3y^2}$ Are the next computations correct?
$$
0\leq \left|\frac{x^5-\sin(y^3)}{4x^4+3y^2}\right|
\leq\frac{|x^5|}{4x^4}+\frac{|y^3|}{3y^2}
=\frac{|x|}{4}+\frac{|y|}{3}
$$
Then, by squeeze theorem, the limit is zero.
 [/INST] $$0\leq |\frac{x^5-\sin(y^3)}{4x^4+3y^2}|\leq\frac{|x^5|}{4x^4+3y^2}+\frac{|y^3|}{3y^2+4x^4} \leqslant\frac{|x|}{4}+\frac{|y|}{3} \leqslant \frac{|x|+|y|}{3}=\frac{\sqrt{x^2}+\sqrt{y^2}}{3}\leqslant \frac{2(\sqrt{x^2+y^2})}{3}=\frac{2 \delta}{3}$$
If you take $\delta=\frac{3 \epsilon}{2}$ you have the $\epsilon-\delta$ proof of your limit.
Also as for a second solution:
$$0\leq |\frac{x^5-\sin(y^3)}{4x^4+3y^2}|\leq\frac{|x^5|}{4x^4+3y^2}+\frac{|y^3|}{3y^2+4x^4}$$
Compute the limits $$ \lim_{(x,y) \rightarrow (0,0)}\frac{|x^5|}{4x^4+3y^2}$$ $$\lim_{(x,y) \rightarrow (0,0)}\frac{|y^3|}{3y^2+4x^4}$$
by taking polar coordinates:
$$x=r\cos{\theta}$$ $$y=r\sin{\theta}$$
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