$$a_n\ge0, l\ge0$$If $$\lim_{n\rightarrow\infty}a_n=l$$ Then $$\lim_{n\rightarrow\infty}a^2_n=l^2$$
Proving $\lim_{n\rightarrow\infty}a_n=l\Rightarrow \lim_{n\rightarrow\infty}a^2_n=l^2$
-
4Your notation is wrong. You could write $\lim_{n \to \infty} a_n = \ell$, or you could write $a_n \to \ell$ as $n \to \infty$. – 2012-11-14
-
0... or sometimes $a_n {{\longrightarrow} \atop {n \to \infty}} \ell$ – 2012-11-14
-
2But don't mix $\lim_{n \to \infty}$ and $\to \ell$. – 2012-11-14
-
1$f(x)=x^2$ is continuous. QED. – 2012-11-14
-
0yuvalz:l Do you plan to correct the $\lim\ldots\to\ldots$ misprints in the title and in the text of the question? – 2012-11-14
-
0Oh, I didn't seem to understand what the problem was till now. Right on it. – 2012-11-14
3 Answers
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 $\lvert a_n^2 - l^2\rvert \rightarrow 0$.
-
0I don't get what you are saying. Can you please try to clarify? – 2012-11-14
-
0@yuvalz I've edited to include a little more information, let me know if it helps or you would like me to clarify further. – 2012-11-14
-
0Why rely on some algebraic specificities of the problem, while it has a completely general solution? – 2012-11-14
-
0Can 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
? – 2012-11-14 -
0@yuvalz, what you should do is fix an $\epsilon$, choose $N_0$ large enough so that both $\lvert (a_n - l) \rvert ^2$ and $\lvert 2l(a_n - l) \rvert$ are less than $\epsilon$. Which you can do because $a_n - l \rightarrow 0$. – 2012-11-14
-
0I'm sorry, I'm not a native english speaker. What do you mean by fix an $\epsilon$? – 2012-11-14
-
0@yuvalz choose any small number, and call it $\epsilon$. Then choose $N_0$ large so that for $n\geq N_0$ we have $\lvert(a_n - l)\rvert^2$ and $\lvert 2l(a_n - l)\rvert$ are both less than $\epsilon$. Then, since we do this for any small number $\epsilon$, we get that $a_n^2 - l^2 \rightarrow 0$. – 2012-11-14
-
0@DevenWare Ok. Can you explain why replacing the $a_n$ with $l$ is invalid in this case? – 2012-11-14
-
0@yuvalz I'm not quite sure what you mean. We don't ever replace $a_n$ with $l$, we just say that eventually $a_n$ is very close to $l$. – 2012-11-14
-
0@DevenWare I've been told that $\lim_{n\rightarrow\infty}a_n\rightarrow l$. Why can't I use it to say $a_n^2-l^2=l^2-l^2$? – 2012-11-14
-
2@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. – 2012-11-14
-
0@DevenWare Can I just ask of you one last thing? Can you please complete your proof, so I'll see how it should look? Many thanks. – 2012-11-14
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 $\lim_{n \to \infty} a_n^2 = l^2$.
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.