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If we have a real sequence $\left|a_n\right|$ such that $\lim_{n \rightarrow \infty} a_n = a$, how do we prove (by an $\epsilon - N$ argument) that $\left|a_n\right|$ such that $\lim_{n \rightarrow \infty} a_{n}^{2} = a^2$?

I know you can use algebra to do to the following:

$$\left|a_n^2 - a^2\right| =\left|(a_n - a)(a_n + a)\right|$$

Where I feel like you can use the implication that $\lim_{n \rightarrow \infty} a_n = a$ to show that $(a_n-a) < a$ or something.

What's the proper way to go about this?

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  • $\begingroup$ As $a_n \to a$ we get $(a_n - a) \to 0$ and $(a_n + a) \to 2a$ therefore the right hand side converges to 0, therefore the left hand side converges to 0. $\endgroup$ – Krijn Sep 20 '15 at 7:06
  • $\begingroup$ Where did you get $(a_n+a)\to 2a$ from? $\endgroup$ – galois Sep 20 '15 at 22:10
  • $\begingroup$ Well, $a_n \to a$? $\endgroup$ – Krijn Sep 20 '15 at 22:54
  • $\begingroup$ Ah, I see. But why are we showing it goes to infinity? or are we just showing that 0 < $\epsilon$? $\endgroup$ – galois Sep 21 '15 at 1:21
  • $\begingroup$ We are not showing that it goes to infinity, we are showing that it is bounded by $2a$, which means that if we multiply it with something that goes to $0$, we get $0$. $\endgroup$ – Krijn Sep 21 '15 at 21:22
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Hint

A convergent sequence is bounded. So you can also bound $\vert a+a_n\vert$.

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Notice, we know $$\lim_{n\to \infty}a_n=a$$ Since the above limit exists as $n\to \infty$ so we can separate the limits in form of product as follows

$$\lim_{n\to \infty}a_n^2=\lim_{n\to \infty}(a_n\cdot a_n)$$$$=\lim_{n\to \infty}(a_n)\cdot \lim_{n\to \infty}(a_n)$$ $$=a\cdot a=a^2$$

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  • $\begingroup$ This is an overkill. You need to prove $\lim (x_ny_n) = \lim x_n \lim y_n $ first. $\endgroup$ – user99914 Sep 20 '15 at 11:39
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If for every $\varepsilon > 0$ there is some $N \geq 1$ such that $|a_{n}-a| < \varepsilon$, then, taking any $\varepsilon > 0$, there is some $N \geq 1$ such that $$ |a_{n}^{2}-a^{2}| = |a_{n}-a||a_{n}+a| < 2|a|\varepsilon+\varepsilon^{2} $$ for all $n \geq N$.

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  • $\begingroup$ I don't understand how you came to the step $\dots < 2\left|a\right|\epsilon + \epsilon^2$. What happened? $\endgroup$ – galois Sep 20 '15 at 22:11
  • $\begingroup$ Triangle inequality. $\endgroup$ – Megadeth Sep 21 '15 at 2:17

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