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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

2 Answers 2

up vote 4 down vote accepted

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

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