Find $\lim_{(x,y) \to (0,0)}\frac{x^3\sin(x+y)}{x^2+y^2}$ Find the limit $$\lim_{(x,y) \to (0,0)}\frac{x^3\sin(x+y)}{x^2+y^2}.$$
How exactly can I do this? Thanks.

Using $|\sin z|\leq 1$ we find that the absolute value of your function is not greater than
$$
\frac{|x^3|}{x^2+y^2}\leq \frac{|x^3|}{x^2}=|x|.
$$
This is first when $x\neq 0$. Then observe that $|f(x,y)|\leq |x|$ also holds when $x=0$.
Can you conclude?
