Solving a limit involving 2 variables I'm trying to solve the following limit:
$$\lim_{(x,y)\to(2,1)} \frac{xy(x-2)^{5/3}(y^2-1)^{2/3}}{|y|(x-2)^2+x^2(y-1)^2}$$
I already know that if it exists the limit is $0$ since approaching $y \to 1$ while fixing $x=2$ yields $0$; but I'm unable to give a $\delta$-$\epsilon$ proof for this. Any ideas?
 A: Try writing
$$
\lim_{(x,y)\to(2,1)} \frac{(x-2)^{5/3}(y^2-1)^{2/3}}{(x-2)^2 + (y-1)^2}
$$
with the substitution $x\mapsto(u+2)$ and $y\mapsto(v+1)$ as
$$
\lim_{(u,v)\to(0,0)} \frac{u^{5/3}v^{2/3}(v+2)^{2/3}}{u^2+v^2}
$$
we know that
$$u^2\le u^2+v^2\tag{1}
$$
and that
$$
2uv\le u^2+v^2\tag{2}
$$
Multiply $(1)$ to the $1/2$ power times $(2)$ to the $2/3$ power to get
$$
2^{2/3}u^{5/3}v^{2/3}\le (u^2+v^2)^{7/6}
$$
So the whole fraction is $\le(u^2+v^2)^{1/6}$ and that $\to0$.
A: HINT:
This is ugly looking, so how about we sandwich a few things to make it prettier? For example, if I just go ahead and limit myself to never take a $\delta > \frac{1}{2}$, then I can say that $1.5 < x < 2.5$ and $.5 < y < 1.5$.
With these, you can sandwich away all the x's and y's that aren't in some expression going to zero. Then you're left with the much easier $\dfrac{(x-2)^{5/3}(y^2-1)^{2/3}}{(x-2)^2 + (y-1)^2}$ with a few constants here and there that I have left out (potentially different for the upper and lower bounds).
