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.
 A: Note that when it comes to $(x,y)\to(0,0)$ we may use that $|x|\leq\sqrt{x^2+y^2}=:r$ and similarly $|y|\leq r$. It follows that $|\sin(x^2y)|\leq |x^2y|\leq r^3$ and therefore
$$\left|{\sin(x^2y)\over x^2+y^2}\right|\leq{r^3\over r^2}=r\ .$$
Assume now that an $\epsilon>0$ is given. If $\bigl\|(x,y)-(0,0)\bigr\|=r<\delta$, where $\delta:=\epsilon$, then the expression in question is $<\epsilon$. This shows that $\delta:=\epsilon$ is o.k.
For the second problem you have to use that$$\bigl|\tan s\bigr|\leq {4\over\pi}\>|s|\qquad\bigl(|s|<\pi/4\bigr)$$
and then argue similarly. You'll have to include a "buffering factor" in the resulting $\delta$.
A: 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 $0\le x\le\frac\pi4$
$$
\tan(x)\le\frac4\pi x
$$
A: For the first one try
$$\frac{|\sin x^2y|}{x^2+y^2}\leq \frac{|x^2y|}{x^2+y^2}=\frac{|y|}{1+\left(\frac{y}{x}\right)^2}\leq |y|$$
A: $\frac{|\sin x^2y|}{x^2+y^2}\leq \frac{x^2 |y|}{2|x| |y|}=\frac{|x|}{2}\to 0$.
No $r$'s, no epsilons, no deltas, no buffering factors.
