Proof - limit of a sequence

**loui1410** · September 19th 2012, 04:39 AM

Prove, by definition, that if $\lim_{n\to \infty} a_n = L$ then $\lim_{n\to \infty} a_n^2 = L^2$ .

It should be a basic proof but I'm somehow stuck. I was obviously thinking:

$\mid a_n^2 - L^2 \mid = \mid a_n - L \mid \cdot \mid a_n + L\mid$

but I'm not sure how to proceed.
Any help would be appreciated.

**emakarov** · September 19th 2012, 04:44 AM

$|a_n-L|$ is small and $|a_n+L|$ is close to 2L and therefore cannot be large.

**loui1410** · September 19th 2012, 04:51 AM

I don't understand...

**emakarov** · September 19th 2012, 05:00 AM

You are given an ε > 0 and you need to make |aₙ - L|⋅|aₙ + L| < ε by choosing n. You can make |aₙ - L| arbitrarily small by choosing n big enough. Also, you can make |aₙ + L| arbitrarily close to 2L by choosing n big enough. In particular, you can make |aₙ + L| < 3L for some n and onward. Assuming you have done this, how small do you need to make |aₙ - L| so that |aₙ - L|⋅|aₙ + L| becomes smaller than ε?

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