<s>[INST] Prove the limit of a function with two variables is zero I have to prove as homework that
$$\lim_{(x,y)\to (0,0)} \frac{\sin\left(x^2y\right)}{x^2+y^2}=0$$
and that
$$\lim_{(x,y)\to (0,0)} \frac{\tan\left(x^3\right)}{x^2+y^2}=0.$$
For the second limit here's what I did:
let $\epsilon >0$ we want $\delta >0$ such that if $\|(x,y)\|<\delta$ then $|tan(x^3)/(x^2+y^2)|<\epsilon$
So suppose $\|(x,y)\|<\delta$, $|tan(x^3)/(x^2+y^2)|= |tan(x^3)/(x^3)||(x^3)/(x^2+y^2)|< |tan(x^3)/(x^3)||x|=|sin(x^3)/(x^3)||1/cos(x^3)||x|<|x|/|cos(x^3)|$
For the first limit I don't know how to begin.
 [/INST] Hint: Use the fact that $\sin(x)$ and $\tan(x)$ are odd to reduce to the case that $x,y\ge0$. Then
$$
x^2\le x^2+y^2\quad\text{and}\quad y\le\left(x^2+y^2\right)^{1/2}
$$
therefore,
$$
\frac{x^2y}{x^2+y^2}\le\left(x^2+y^2\right)^{1/2}
$$
Then use the fact that for $x\ge0$, $\sin(x)\le x$.

For the second inequality, you can use the fact that $\tan(x)$ is convex on $\left[0,\frac\pi2\right)$and $\tan\left(\frac\pi4\right)=1$ to see that for $0\le x\le\frac\pi4$
$$
\tan(x)\le\frac4\pi x
$$
</s>