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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2Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – 2012-09-26
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1Please tell us what you have already tried, and what difficulty you have experienced. – 2012-09-26
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0I have tried to find some \delta, but couldn't find – 2012-09-26
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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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0So in order to prove with the $\delta$ method, is it OK to just repeat what you showed, and pick a value randomly, and set it to Limit, and prove that other values have the different limits, no matter what $\delta$ I choose? – 2012-09-26
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0It sounds like you've probably got it figured out. Let me put it another way to be sure: Find some $\alpha,\beta$ such that $\cfrac{2\alpha}{1+\alpha^2}\neq\cfrac{2\beta}{1+\beta^2}$, and set $\varepsilon$ less than half of $\left|\cfrac{2\alpha}{1+\alpha^2}-\cfrac{2\beta}{1+\beta^2}\right|$. Suppose by way of contradiction that the limit exists, say $L$. Show that for any $\delta>0$, there is some $x\neq 0$ such that both $(x,\alpha x^2)$, $(x,\beta x^2)$ are within $\delta$ of $(0,0)$. By triangle inequality, the function values are within $2\varepsilon$ of each other. Contradiction. – 2012-09-26
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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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0No. I haven't used delta-eplison to prove some limit doesn't exist. – 2012-09-26
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0@FrankXu: what happens if $x=y?$ – 2012-09-26
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0How can I do this kind of problem? Thank you! – 2012-09-26
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0Yeah, I know how to do this problem by setting $x = y^2$, but how to do this with $\eplison$ and $\delta$? – 2012-09-26
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0@FrankXu: you are backwards in my hints. If you put $x^2=y$ into your original problem, what happens? – 2012-09-26
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0The expression equals to 1 – 2012-09-26
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0Then what should I do? I'm not familiar with this kind of proof. Thank you! – 2012-09-26
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0@FrankXu: The point is that there are points close to $(0,0)$ where the expression is far from $0$. There are others (where $y \ll x$ for example) where it is close to $0$. The specific $\epsilon - \delta$ proof is I give you $\epsilon = \frac 14$. Now you tell me what $\delta$ I need to be close to $(0,0)$ so the function is less than $\frac 14$. Whatever you are given, you can find a point where the function is $\frac 12$ so the limit fails. – 2012-09-26
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0But how could you know that 0 is the limit(if it exists)? It could be anything. So, in order to prove this, am I supposed to find some values of $x$ and $y$ so that limit is 0, and disprove this with other values? That would be similar to the solution below. – 2012-09-26
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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