Proving from first principles that a(n)^2 -> 4 if a(n) -> 2.

[Tommy941]

Homework Statement

Let a_n → 2. Prove from first principles (i.e. give a direct ε-N proof) that a_n² → 4.

Homework Equations

The Attempt at a Solution

I have tried considering |a_n-2|² and considering that |a_n²-4| = |(a_n+2)(a_n-2)| but I could not get either of these methods to work. Would someone be able to point me in the right direction?
I was thinking maybe one of the sequences is a subsequence of the other?

I was also wondering, is this true generally that if a sequence tends to L then the sequence squared will tend to L²?

Thanks,
Tommy

 

[HallsofIvy]

You are given that, for \epsilon> 0, there exist N such that is n> N, |a_n-2|< \epsilon (since a_n goes to 2).

Now, you want |a_n^2- 4|= |a_n+ 2||a_n- 2|< \epsilon. If you can find an upper bound, A, on |a_n+ 2| then you can say that |a_n+ 2||a_n- 2|< A|a_n- 2| so that if A|a_n- 2|< \epsilon, which is the same as |x_n- 2|< \epsilon/A, then |a_n+ 2||a_n- 2|< A|a_n- 2|< \epsilon.

Now, if a_n is close to 2, say, |a_n- 2|< 1, what can you say about |a_n+ 2|?

 

[RUber]

Start with something like :
There exists a delta such that if |an-2|<delta, then (an-2)^2 < epsilon.
Then notice that (an+2) = (an-2)+4.
I think those are the main techniques in this proof.

 

[Tommy941]

Thanks HallsofIvyy I think I will be able to do it now. Is it O.K for me to say that for suitably large n, |a_n-2|<ε<1 ⇒ |a_n+2|<3 so 3 can be considered an upper bound on a_n when n is suitably large??

 

[HallsofIvy]

Yes, that will work.