$\lim_{(x,y)\rightarrow (0,0)}\frac{x^5-\sin(y^3)}{4x^4+3y^2}$

Question

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.

Answer 1

$$0\leq |\frac{x^5-\sin(y^3)}{4x^4+3y^2}|\leq\frac{|x^5|}{4x^4+3y^2}+\frac{|y^3|}{3y^2+4x^4} \leqslant\frac{|x|}{4}+\frac{|y|}{3} \leqslant \frac{|x|+|y|}{3}=\frac{\sqrt{x^2}+\sqrt{y^2}}{3}\leqslant \frac{2(\sqrt{x^2+y^2})}{3}=\frac{2 \delta}{3}$$

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:

$$x=r\cos{\theta}$$ $$y=r\sin{\theta}$$