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

Tags:
1. Feb 28, 2015

### Tommy941

1. The problem statement, all variables and given/known data
Let an → 2. Prove from first principles (i.e. give a direct ε-N proof) that an2 → 4.

2. Relevant equations

3. The attempt at a solution
I have tried considering |an-2|2 and considering that |an2-4| = |(an+2)(an-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

2. Feb 28, 2015

### 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 $|a_n+ 2|$?

Last edited by a moderator: Feb 28, 2015
3. Feb 28, 2015

### RUber

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.

4. Feb 28, 2015

### 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, |an-2|<ε<1 ⇒ |an+2|<3 so 3 can be considered an upper bound on an when n is suitably large??

5. Feb 28, 2015

### HallsofIvy

Yes, that will work.