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I was wondering if the following is true...

If $(a_{n})$ is a sequence of positive terms converging to $a$ and $(b_{n})$ is a real sequence converging to $b$, then the sequence $(a_{n}^{b_{n}})$ converges to $a^{b}$.

This is what I did

$$ a_{n}^{b_{n}} = e^{b_{n}\ln{a_{n}}}$$

Now as the logarithm function is continuous, we have, for a fixed $m$

$$ \lim\limits_{n\to \infty}a_{n}^{b_{m}} = e^{b_{m}\ln{a}} $$

Now as exponential is continuous, we have our result.

Is every step here correct? I am skeptical because this is an important result and the proof is easy (if it is correct) and yet it is not found in real analysis books.

EDIT: We have $(1 + 1/n) \to 1$ and $n\to \infty$ and $(1 + 1/n)^{n} \to e$. Does this qualify as a counterexample? Is the result only true for finite $a$ and $b$?

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2 Answers 2

up vote 4 down vote accepted

The proof is not quite correct. You can't decide which variable to keep fixed and which one to compute a limit on (as your edit shows, one can obtain lots of nonsense by doing so). Instead, simply argue by continuity of $\ln$ that $\ln (a_n)$ converges to $\ln (a)$ (assuming here $a\ne 0$), and then it follows that $b_n\cdot \ln(a_n)$ converges to $b\ln(a)$. Then the computation is complete.

The proof is easy but it uses the continuity of the exponential function and of $\ln$, as well as the fact that $a^b=e^{b\ln a}$. None of these is particularly trivial. Notice that you can define exponentials from first principles, only using the algebraic definitions up to when the exponent is a rational number (all positive). Then you can define arbitrary exponents by continuity. This will produce an elementary a rigorous definition of exponents, including of the exponential function itself. If this is done, then $a^b$ is defined to be the limit $a_n^{b_n}$ where $a_n$ is a sequence of rationals converging to $a$, and $b_n$ is a sequence of rationals converging to $b$. It is then elementary and not too hard to show that $a^b$ is independent of the choice of sequences.

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I would skip the step of disconnecting the indices of the two sequences.

Rather: by continuity of the logarithm, $\ln a_n\rightarrow a$ as $n\rightarrow\infty$. Hence $b_n\ln a_n\rightarrow b\ln a$ as $n\rightarrow\infty$; so, by continuity of the exponential, $a_n^{b_n}\rightarrow a^b$.

Note that you should also explicitly add the assumption that $a>0$.

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