# Simple analysis proof: x^n -> a^n as x -> a

1. Sep 29, 2009

### applegatecz

1. The problem statement, all variables and given/known data
Prove that x^n approaches a^n as x approaches a.

2. Relevant equations

3. The attempt at a solution
I understand the concept here ... need to find a delta>0 for epsilon>0 s.t. |x-a|<delta implies |x^n-a^n|<epsilon. For some reason I can't solve this one. Thanks.

2. Sep 29, 2009

### Dick

You know you can factor x^n-a^n=(x-a)*(x^(n-1)+x^(n-2)*a+ ... + a^(n-1)), right?

3. Sep 29, 2009

### applegatecz

Oh! Somehow I missed that. Thank you very much for your help.

4. Sep 29, 2009

### g_edgar

Dick assumes n is a positive integer, which you did not tell us under "all variables and given/known data" ... so are we to assume n is NOT know to be a positive integer?

5. Oct 2, 2009

### economist1985

still, how do you show that (x^(n-1)+x^(n-2)*a+ ... + a^(n-1)) is < epsilon?

6. Oct 2, 2009

### HallsofIvy

Staff Emeritus
You don't, and no one has claimed it is. You want $|x^n- a^n|= |x- a||x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|< \epsilon$ for |x- a| small enough. That is the same as choosing $|x- a|< \epsilon/|x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|$ and you can do that by finding any positive lower bound on $|x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|$ for x close to a.

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