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Let sequences $\{a_n\}$ and $\{b_n\}$ have the property that $\lim_{n \to \infty} [a_n^2+b_n^2]=0$. Prove that $\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=0$.

Proof: Suppose that $\lim_{n \to \infty} [a_n^2+b_n^2]=0$ and without loss of generality assume that$\{a_n\}$ diverges. It follows that there exists a $\epsilon>0\ \implies\sqrt{\epsilon}>0$ then there exists a $N\in\mathbb{N}$ such that if $n\geq N$ then $a_n>\sqrt{\epsilon}\implies a_n^2>\epsilon$. But since $\lim_{n \to \infty} [a_n^2+b_n^2]=0$ it follows that there exists a $N'\in\mathbb{N}$ such that if $n\geq N'$ then $|a_n^2+b_n^2|=a_n^2+b_n^2<\epsilon$ a contradiction. Thus $\{a_n\}\rightarrow a$ and $\{b_n\}\rightarrow b$.

Now we show $\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n= 0$ Suppose not that $\lim_{n \to \infty} a_n\neq \lim_{n \to \infty} b_n\neq 0$. Then it follows that $a\neq b$ and $a,b\neq 0$. But that's a contradiction since $\lim_{n \to \infty} [a_n^2+b_n^2]=a^2+b^2=0$

Would this be correct?

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This proof isn't really sufficient: The limit law that you use assumes that $a_n^2$ and $b_n^2$ are already both convergent sequences, which you're not given. Also, I'm not sure where exactly $a$ and $b$ come from - are these the limits of $a_n$ and $b_n$ (which we also don't know exist)?


A better way would be something like this:

Suppose that $a_n$ did not converge to $0$. Then there exists an $\epsilon > 0$ such that for arbitrarily large $n$, we have $$|a_n| > \epsilon$$

Now conclude that $a_n^2$ is large for arbitrarily large $n$, and use it to contradict that $a_n^2 + b_n^2 \to 0$.

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  • $\begingroup$ ok i fixed it. would it be right now? $\endgroup$ – user60887 Nov 17 '13 at 2:20
  • $\begingroup$ @user60887 Your quantifiers are out of order. There exists some $\epsilon$; the statement doesn't hold for all $\epsilon > 0$. Otherwise, it's fine. $\endgroup$ – user61527 Nov 17 '13 at 2:21
  • $\begingroup$ oh ok ill fix it. $\endgroup$ – user60887 Nov 17 '13 at 2:22
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First we can note that $0 \leq a_{n}^{2} \leq a_{n}^{2} + b_{n}^{2}$ and then by Sandwich theorem $\lim_{n \to \infty}a_{n}^{2} = 0$. Similarly $\lim_{n \to \infty}b_{n}^{2} = 0$. Next let $\epsilon > 0$ be an arbitrarily preassigned number. Then since $a_{n}^{2} \to 0$ as $n \to \infty$ it follows that there is a positive integer $m$ such that $a_{n}^{2} < \epsilon^{2}$ whenever $n > m$. It follows by taking square roots that $|a_{n}| < \epsilon$ when $n > m$. So $\lim_{n \to \infty}a_{n} = 0$. The result for $b_{n}$ follows in a similar manner.

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  • $\begingroup$ very nice. I like this way. $\endgroup$ – user60887 Nov 17 '13 at 4:37
  • $\begingroup$ To user60887: Yes you can use the proofs based on Sandwich theorem for your question. I find it a very useful technique. I hope you are aware of this theorem (also known as Squeeze theorem) $\endgroup$ – Paramanand Singh Nov 17 '13 at 4:38
  • $\begingroup$ yes I am but usually I don't use it too often. $\endgroup$ – user60887 Nov 17 '13 at 4:40

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