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We are given two real sequences $(a_n)_{n\in \mathbb{N}}$ and $(b_n)_{n\in \mathbb{N}}$ such that $a_n$ > 0 and $\lim_{n\to\infty}a_n = a > 0$, as well as $\lim_{n\to\infty}b_n = b$.

We would now like to show that $\lim_{n\to\infty}a_n^{b_n} = a^b$, that is

$\forall\varepsilon > 0:\exists N_\varepsilon\in\mathbb{N}:\forall n \ge N_\varepsilon: | a_n^{b_n} - a^b | < \varepsilon$.

I'm sure this can't be that hard but so far this has me stumped.

I tried to use the triangle inequality to reduce the above expression to the limit conditions for $a_n$ and $b_n$ respectively,

$|a_n^{b_n} - a^b| \le |a_n^{b_n} - a_n^b| + |a_n^{b} - a^b|$.

But since $b$ is generally real in the second term I cannot use the fact that the exponent "pulls through" the limit, i.e. $\lim_{n\to\infty}a_n^k = a^k$, for $k\in\mathbb{N}$, or can I?

In the first term I tried to estimate an upper bound using the fact that $(b_n)$ converges and hence the sequence must be bounded, but since we only know that $a_n$ > 0 and not $a_n \ge 1$, this gets me nowhere.

Any guidance in the right direction would be very much appreciated!

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$a_n^{b_n}= e^{b_n \log a_n} \to e^{b \log a}=a^b$, since the functions $e^x$ and $ \log x$ are continuous.

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  • $\begingroup$ This makes sense, of course, but it is not really in the spirit of the problem. The point is to prove it using the $\varepsilon$-$N_\varepsilon$ criterion, maybe I should have added this to the question from the start. $\endgroup$ – hemmlmann Nov 16 '17 at 9:50

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