When and for what values of x does lim anxn exist?

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Homework Statement
Let an be a sequence of real numbers. For what values of x does lim anxn exist?

The attempt at a solution
Let us suppose that lim anxn exist and is equal to b. What can we say about x? Hmm...there is a monotonic subsequence that converges to b, say a_{k_n}x^{k_n}. If this is an increasing sequence, we have that

<br /> a_{k_n}x^{k_n} \le a_{k_{n+1}}x^{k_{n+1}}<br />

or equivalently

<br /> \frac{a_{k_n}}{a_{k_{n+1}}} \le x^{k_{n+1} - k_n}<br />

Unfortunately I don't get an inequality in terms of x alone. How do I proceed from here? Perhaps I need a further assumption, like k_{n+1} - k_n = 1 for all n?
 
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its going to depend on your a_n... if a_n =0 for all n, it exists for every x
 
sorry, actually I misread sequence

if the limit exists, and is b say, then for any e>0 you can choose N such that for all n>N
|a_nx^n -b|&lt;e

you could then think about the beahviour of x when x>|1| and otherwise

though as mentioned previously it will depend on teh sequnce a_n , is it for any a_n or are there any contraints on the a_n ?
 
There are no constraints on an. I do know of a couple that might shed some light on this general situation. As you previously mentioned, if the an are all 0 (or are eventually all 0), then the limit exists for all n. If the an are eventually all some nonzero constant, then we know the limit exists for all |x| < 1. Now I believe that if c = lim |an/an+1|, then the limit exists for all |x| < c. This is something I have yet to prove though.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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