As the topic, how to prove by $\epsilon$-$\delta $ approach $\lim_{(x,y)\rightarrow (0,0)}\frac {x^n-y^n}{|x|+|y|}$ exists for $n\in \mathbb{N}$ and $n>1$
proving by $\epsilon$-$\delta $ approach that $\lim_{(x,y)\rightarrow (0,0)}\frac {x^n-y^n}{|x|+|y|}$exists for $n\in \mathbb{N}$ and $n>1$
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2(As you know very well:) What did you try? Where are you stuck? What similar problems can you solve? – 2012-03-04
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0I have tried the simpliest case when n=2 and trying to show $||(x,y)||<\delta,\implies |\frac {x^2-y^2}{|x|+|y|}|< \epsilon$ but still cannot set up the inequality – 2012-03-04
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0For 2D, note that $\|x^2-y^2\|=\|(x+y)(x-y)\|$ and that |x|+|y|>=|x+y|. So, $|x^2-y^2|/|x+y| \le |x-y| \le 2\delta$ – 2012-03-04
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2TeXtip: write `$\epsilon$-$\delta$` and not `$\epsilon-\delta$`, so as to get $\epsilon$-$\delta$ and not $\epsilon-\delta$: you do not want to substract anything... – 2012-03-04
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1@MarianoSuárez-Alvarez : I was about to say the same thing. Often one sees people putting a hyphen where a minus sign should be; this is the opposite mistake. The difference is not only that a minus sign is longer, but also that it has a space before and after it when it's a binary operation (e.g. $3 - 5$) but not when it's a unary operation (e.g. $-5$). Apparently some people don't notice things like this. – 2012-03-04
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0@Mathematics, I think it should be interesting you verify why the result doesn't holds for $n=1$. – 2012-03-05
4 Answers
Let $\epsilon>0$. There is an $\delta>0$ such that $\xi^n\leq \epsilon\xi$ for all $\xi\in[0,\delta)$. Then if $(x,y)\in(-\delta,\delta)\times(-\delta,\delta)$ and $(x,y)\neq0$, we have $$\left|\frac{x^n-y^n}{|x|+|y|}\right|\leq\frac{|x|^n+|y|^n}{|x|+|y|}\leq\frac{\epsilon(|x|+|y|)}{|x|+|y|}=\epsilon.$$ We win.
P.S. Mathematics wants to know how to prove the existence of $\delta$. One can proceed like Neal suggests, or various variations of that idea. A simpler approach is the following. If $0\leq\xi\leq\min\{\epsilon,1\}$ then $$0\leq\xi^n=\xi\cdot\xi\cdot\xi^{n-2}\leq\epsilon\cdot\xi\cdot1=\epsilon\xi$$ because $0\leq\xi^{n-2}\leq1$. This means that we can take $\delta=\min\{\epsilon,1\}$. :)
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0nice solution :) +1 because, I think this solution is a prime example of the benefit of thinking twice before you start writing down a solution! – 2012-03-04
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2how do you know $\xi^n\leq \epsilon\xi$? – 2012-03-04
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0Try thinking about it upside-down. If $\epsilon$ and $n$ are fixed, there is some number $x$ which has $x^n = \epsilon^{-1} x$. Then for every $\xi^{-1} \geq x$, $\xi^{-n} \geq \epsilon^{-1} \xi^{-1}$. Now take inverses of both sides, so the inequality reverses: for each $\xi \leq x^{-1}$, $\xi^n\leq \epsilon\xi$. Put $\delta = x$ and you're done. – 2012-03-04
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0Very slick! Why did this answer get voted down? – 2012-03-04
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3I *am* curious about the reason for the downvotes :) – 2012-03-05
Hint 2: Working from what you have (as per your comment on user22705's answer), observe that $$\begin{eqnarray}(|x|+|y|)^2&=&x^2+y^2+2|x||y|\\ &\leq& x^2+y^2+2x^2+2y^2\\ &=&3(x^2+y^2)\end{eqnarray}$$
A HINT is to rewrite the numerator using the following identity: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+….+b^{n-1})$ then use the triangle inequatlity.
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0i have tried that and got $\frac {x^n-y^n}{|x|+|y|}<|x|+|y|$but i can't garantee that $|x|+|y|\le\(x^2+y^2)^{1\over 2}$ or something similar – 2012-03-04
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0try the following $|x^n-y^n|=|(x-y)||(x^{n-1}+x^{n-2}y+….+y^{n-1})|<(|x|+|y|)|(x^{n-1}+x^{n-2}y+….+y^{n-1})|<(|x|+|y|)(|x|^{n-1}+|x^{n-2}||y|+….+|y^{n-1}|)|$ – 2012-03-04
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0now the numerator and the denominator should cancel. the expression that remains is much easier to "work" with. do you see? – 2012-03-04
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2Mathematics: Coming back to your first comment: surely you can compare $|x|$ and $(x^2+y^2)^{1/2}$! And $|y|$ and $(x^2+y^2)^{1/2}$. Thus bounding $|x|+|y|$ by a multiple of $(x^2+y^2)^{1/2}$ should be doable, no? – 2012-03-04
You may use that
$$\left|\frac{x^n-y^n}{|x|+|y|}\right|\leq \frac{|x|^n-|y|^n}{|x|+|y|}\leq \frac{|x|}{|x|+|y|}|x|^{n-1}+\frac{|y|}{|x|+|y|}|y|^{n-1}\leq|x|^{n-1}+|y|^{n-1}.$$
Since you impose $x^2+y^2< \delta \leq 1$ you have $|x|, |y|<1\Rightarrow |x|^{n-1}<|x|,\ |y|^{n-1}<|y|.$
Then you have
$|x|^{n-1}+|y|^{n-1}<|x|+|y|\leq 2\sqrt{x^2+y^2}$.
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0hey,we didn't know $\delta\le 1$,it should be any $\delta >0$ – 2012-03-05
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0The $\epsilon$ is arbitrary but the $\delta_\epsilon$ should be only positive. However you may also suppose $\epsilon$ small according to your needs because the $\delta$ for an small $\epsilon$ will suit for a great as well. – 2012-03-05
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0what if i choose $\delta =2$ then your last inequality doesn't holds – 2012-03-05
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0The definition of limit will be more important than this naive example: We say that $$lim_{(x,y)\to (a,b)}f(x,y)=L$$ when for all $\epsilon >0$ there is a $\delta=\delta_{\epsilon}>0$ such that $$||(x,y)-(a,b)||<\delta \Rightarrow ||f(x,y)-L||<\epsilon$$ – 2012-03-05