Prove that the limit $\lim\limits_{(x,y)\to(0,0)}\frac{2x^2 y}{ x^4 + y^2}$ when doesn't exist with an $\varepsilon$-$\delta$ argument.
Prove limit doesn't exist using $\varepsilon$ and $\delta$
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0That is what you would do if you were trying to find the limit, or to prove that the limit exists. Here, you need to do the opposite - show that no matter what $\delta$ is, you can't force the expression fall within $\epsilon$ of some value by choosing $x$ and $y$ within $\delta$ of 0. – 2012-09-26
2 Answers
Upshot: The idea here is that you want to choose two different paths to the origin--which will allow you to rewrite in terms of only one variable--and use delta-epsilon to prove what the limits along those paths are. To show that the more general limit (that is, not restricted to a particular path) fails to exist, you need to pick your two paths so that the limits along those paths are not the same. The key result you'll need to know (and use) here is that a real-valued function cannot converge to distinct limits as we approach a given point.
A nice way to go here is to choose functions $y=f(x)$ and $y=g(x)$ for your two paths such that $f(0)=g(0)=0$, and such that you get some nice cancellation to allow easy limit evaluation (and consequently, easier delta-epsilon work). In particular, it'd be excellent if we could either get the exponents of $x$ and $y$ to match in the denominator or get rid of one of $x,y$ altogether. For example, if we move along the curve $y=\alpha x^2$ for some constant $\alpha$, then $(x,y)\to(0,0)$ if and only if $x\to 0$. Then substituting $y=\alpha x^2$, we have $\lim_{(x,\alpha x^2)\to(0,0)}\frac{2x^2\cdot\alpha x^2}{x^4+(\alpha x^2)^2}=\lim_{x\to 0}\frac{2\alpha x^4}{(1+\alpha^2)x^4}=\lim_{x\to 0}\frac{2\alpha}{1+\alpha^2}=\frac{2\alpha}{1+\alpha^2}.$ This works for any alpha, and hopefully you can see that different $\alpha$ can result in different limits. Thus, the general limit cannot exist. For an altogether different example, we could approach along the line $x=0$.
What I've shown above isn't the rigorous delta-epsilon proof you need, but the fact that all variables disappeared in the process of evaluating the limit should indicate just how trivial the delta-epsilon proof will actually be....
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0For simplicity, you may as well take $\alpha=1$, $\beta=0$. – 2012-09-26
Hint: define $z=x^2$. Have you seen a problem with $\lim\limits_{(x,y)\to(0,0)}\frac {xy}{x^2+y^2}$?
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0@FrankXu: if you can find $(x,y)$ points arbitrarily close to $(0,0)$ where the function is far (in this case at least $\frac 14$) from $0$, then $0$ is not a limit. My points have the function value $\frac 12$, but there are other points where the value is close to $0$. The definition of $\lim\limits_{(x,y)\to(0,0)}$ is that the function value should be close to the limit everywhere close to the origin. – 2012-09-26