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$$a_n\ge0, l\ge0$$If $$\lim_{n\rightarrow\infty}a_n=l$$ Then $$\lim_{n\rightarrow\infty}a^2_n=l^2$$

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Your notation is wrong. You could write $\lim_{n \to \infty} a_n = \ell$, or you could write $a_n \to \ell$ as $n \to \infty$. – Robert Israel Nov 14 '12 at 5:16
... or sometimes $a_n {{\longrightarrow} \atop {n \to \infty}} \ell$ – Robert Israel Nov 14 '12 at 5:20
But don't mix $\lim_{n \to \infty}$ and $\to \ell$. – Robert Israel Nov 14 '12 at 5:21
$f(x)=x^2$ is continuous. QED. – Gautam Shenoy Nov 14 '12 at 5:25
yuvalz:l Do you plan to correct the $\lim\ldots\to\ldots$ misprints in the title and in the text of the question? – Did Nov 14 '12 at 9:13
up vote 4 down vote accepted

Here is a trick, which will help you solve this problem.

$$a_n^2 - l^2 = (a_n - l)^2 + 2l(a_n - l)$$ using this you should be able to show that $a_n^2 - l^2 \rightarrow 0$.

EDIT: To elaborate further, by the triangle inequality we now know that $$\lvert a_n^2 - l^2\rvert \leq \lvert (a_n - l)^2 \rvert + \lvert 2l(a_n-l)\rvert \hspace{10mm} (1)$$ So we just need to show that each of the two terms on the right get small as $n$ gets large. You should be able to make both of them small by using the fact that $a_n \rightarrow l$.

FINAL EDIT: To see a complete proof. Fix $\epsilon > 0$. Choose $N_0$ so that for $n\geq N_0$, $$\lvert a_n - l\rvert \leq \sqrt{\epsilon};$$ and $N_1$ large so that for $n\geq N_1$, $$\lvert a_n - l \rvert \leq \frac{1}{\lvert 2 l \rvert} \epsilon.$$Then if $n\geq\max(N_0,N_1)$, the inequality (1) above gives us $$\lvert a_n^2 - l^2 \rvert \leq \epsilon + \epsilon = 2 \epsilon$$ and since $\epsilon$ was arbitrary we conclude that $\lvert a_n^2 - l^2\rvert \rightarrow 0$.

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I don't get what you are saying. Can you please try to clarify? – yuvalz Nov 14 '12 at 5:29
@yuvalz I've edited to include a little more information, let me know if it helps or you would like me to clarify further. – Deven Ware Nov 14 '12 at 5:32
Why rely on some algebraic specificities of the problem, while it has a completely general solution? – Did Nov 14 '12 at 6:34
Can I simply say, basen on: $a_n\rightarrow l$ for $n> N_0$. That for $\epsilon>0$: $|a_n^2-l^2|<\epsilon \Leftrightarrow -\epsilon<l^2-l^2=0<\epsilon$ ? – yuvalz Nov 14 '12 at 7:13
@yuvalz $a_n \rightarrow l$ this is true, but this means that $a_n$ is eventually very close to $l$, not that $a_n$ is eventually $l$ itself, which your equality implies. I advise you to read up a little on limits and hopefully this will make more sense. – Deven Ware Nov 14 '12 at 7:42

Since $a_n \to l$, there is a $n_1$ such that $|a_n-l| < 1$ for $n > n_1$.

So, for $n > n_1$, $|a_n^2-l^2| = |(a_n-l)(a_n+l)| < |a_n-l|(2|l|+1) $.

Let $n_2$ be such that $n_2 > n_1$ and $|a_n-l| < \epsilon/(2|l|+1)$ for $n > n_2$. Then $|a_n^2-l^2| < \epsilon$ for $n > n_2$, so $\lim_{n \to \infty} a_n^2 = l^2$.

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Suppose: $$\lim_{n\rightarrow\infty}a_n^2\not\rightarrow l^2$$ Exists an $\epsilon>0$ such that: $$|n^2-l^2|\ge\epsilon \Leftrightarrow l^2-\epsilon\ge n^2 \ge l^2 + \epsilon$$

Left side of the equation is false, because if it were true, $\epsilon<0$. Which is false.

Right side is false because:$$n^2-l^2\ge\epsilon \Leftrightarrow -l^2 \ge \epsilon-n^2 \ge 0$$

Which is again false, because $n,l\ge0$

EDIT: If this is wrong, please let me know why.

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