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daniel_i_l

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## Homework Statement

[tex]a_{n} -> a [/tex] , a>0 and [tex]a_{n}>0[/tex] for all n. Prove that if [tex]b_{n} -> b[/tex] then [tex]{(a_{n})}^{(b_{n})} -> a^b[/tex].

(-> means the limit as n goes to infinity).

## Homework Equations

## The Attempt at a Solution

I first split up a_n into to sub series: [tex]a_{n_{k}}[/tex] and [tex]a_{n_{j}}[/tex] where for all k [tex]a_{n_{k}} >= 1 [/tex]and for all j [tex]0<a_{n_{j}}<1[/tex] . Now, starting with the first series, for all E>0 we can find a number N so that for all k>N [tex] b-E < b_{n_{k}} < b+E [/tex]and so

[tex]{(a_{n_{k}})}^{(b-E)} < {(a_{n_{k}})}^{(b_{n_{k}})} < {(a_{n_{k}})}^{(b+E)} [/tex]. I know that [tex] {(a_{n_{k}})}^{(b+E)} -> a^{(b+E)} [/tex] and so that since E can be as small as we want we should be able to use the sandwich rule to prove that [tex] {(a_{n_{k}})}^{(b_{n_{k}})} -> a^b [/tex] and we can do the same to prove that [tex] {(a_{n_{j}})}^{(b_{n_{j}})} -> a^b [/tex] . But how can I write it in a rigorous way?

Thanks.

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