Limit of $\frac{3xy^2}{x^2+y^2}$ as $(x,y)$ approaches $(0,0)$ Hello Math StackExchange,
I came across this problem in my multivariable calculus class and am wondering if I am taking the correct approach.
$$\lim_{(x,y)\to(0,0)}\frac{3xy^2}{(x^2+y^2)}$$
The discussion below of inequalities excludes the case where $(x,y)=(0,0)$. So I began by restricting $x>0$ so that:
$$ 0\leq \frac{3xy^2}{(x^2+y^2)} \leq \frac{3xy^2+3x^3}{(x^2+y^2)}=3x $$
and proceed to use the Sandwich Theorem to show that $$\lim_{(x,y)\to(0^+,0)}\frac{3xy^2}{(x^2+y^2)}=0$$
Equivalently, when $x<0$:
$$ 0\geq \frac{3xy^2}{(x^2+y^2)} \geq \frac{3xy^2+3x^3}{(x^2+y^2)}=3x $$
and we have in an identical way $$\lim_{(x,y)\to(0^-,0)}\frac{3xy^2}{(x^2+y^2)}=0$$ so we that $$\lim_{(x,y)\to(0,0)}\frac{3xy^2}{(x^2+y^2)}=0$$
Is this approach valid? Any guidance is greatly appreciated. Also note that the question asks specifically to use the Sandwich Theorem and not other methods such as change of basis/coordinates.
 A: The approach is almost valid. You have the case of $x=0$ and $x$ changing signs rapidly near the origin as cases you still need to check. This can be done altogether by using a absolute values so that we have
$$0\le\frac{3|x|y^2}{x^2+y^2}\le\frac{3|x|(x^2+y^2)}{x^2+y^2}=3|x|$$
which gives $0$ as the limit by squeeze theorem.
Alternatively, another good strategy is to try using $r^2=x^2+y^2$, which gives $r\ge|x|$ and $r^2\ge y^2$ so that
$$0\le\frac{3|x|y^2}{x^2+y^2}\le\frac{3r^3}{r^2}=3r$$
which is essentially the same as what you had done, but imo easier to see, as it does not require looking for something in the numerator to cancel with the denominator.
A: Here's a way: 
Observe $$y^2\leq x^2+y^2$$ since $x^2\geq0.$ Therefore $$\frac{y^2}{x^2+y^2}\leq1\implies\frac{3xy^2}{x^2+y^2}\leq3x\leq3|x|$$
Thus $$\lim_{(x,y)\to(0,0)}\frac{3xy^2}{x^2+y^2}\leq\lim_{(x,y)\to(0,0)}3|x|$$
which is equivalent to $$\lim_{(x,y)\to(0,0)}3(-x)\leq\lim_{(x,y)\to(0,0)}\frac{3xy^2}{x^2+y^2}\leq\lim_{(x,y)\to(0,0)}3x$$ 
Can you finish from here? 
