TITLE: Convergence of sequence of real powers
QUESTION [1 upvotes]: Let $\ 0<a_n→0 $ and $\ b_n→b>0 $. Prove $\lim_{n \to \infty}a_n^{b_n}=0 $.
This is what I'd like to do  but I'm not sure how to continue.
Let $\ \epsilon>0 $ we know that there exists N s.t. for every $\ n>N$ $\ |a_n|<\epsilon$ and $\ |b_n-b|< \epsilon$. 
From this we gather $\ (-\epsilon)^{b-\epsilon}<a_n^{b_n}<(\epsilon)^{b+\epsilon}$. 
I'd like to define $\ \epsilon$ as a series $\epsilon _k >0$ s.t. $\lim_{n \to \infty}\epsilon_k=0$. 
I am unsure how to proceed from here.
Would it be sufficient to say:
$\ 0<a_n^{b_n}<(\epsilon_k)^{b+\epsilon_k}$. 
Using limit arithmetic we know that $\ {b+\epsilon_k}→b $ and$\ \epsilon_k→0$
So $\ (\epsilon_k)^{b+\epsilon_k}→0$ and using the squeeze theorem we know that $\lim_{n \to \infty}a_n^{b_n}=0 $. 
Does that suffice?

REPLY [1 votes]: This is a little different from your approach: There exists $N$ such that $0<a_n <1, b_n> b/2$ for $n\ge N.$ For such $n,$ we have
$$\tag 10< a_n^{b_n} < a_n ^{b/2}.$$
(A higher power of a number in $(0,1)$ gives us something smaller.) The function $f(x) = x^{b/2}$ is continuous on $[0,\infty).$ Note that the right side of $(1)$ is $f(a_n).$ Since $a_n \to 0,$ the right side of $(1)$ approaches $f(0)  =0.$ We're done by the squeeze theorem.
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