# 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?

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Giving an epsilon-delta proof of this would masochistic. You are far better off using the established limit laws. – Ragib Zaman Oct 6 '11 at 14:23
Sorry about the poor TeX markup, I'm using my phone to type this. – F M Oct 6 '11 at 14:26

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 $\to0$.

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This was amazing, thanks! – F M Oct 7 '11 at 23:22
@Fernando: in much the same way, you can always get $$x^\alpha y^{1-\alpha}\le\sqrt{x^2+y^2}$$ for $\alpha\in[0,1]$. – robjohn Oct 7 '11 at 23:28

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).

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I was actually asking about how would you solve the limit of $\dfrac{(x-2)^{5/3}(y^2-1)^{2/3}}{(x-2)^2 + (y-1)^2}$; I should have mentioned that I tried out sandwiching non-vanishing terms. – F M Oct 6 '11 at 16:15
could someone explain to me how are the x's and y's that aren't in any expression going to zero eliminated? Are you guys says that the original limit is bound by this simplified limit? thank you – milo Apr 13 '15 at 22:24