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I am working on a proof, which is complete, conditional on the following conjecture being true:

Let $x$ be a positive real number; let $(q_n)_{n=1}^{\infty}$ and $(r_n)_{n=1}^{\infty}$ be convergent sequences of rational numbers (with the limits not necessarily rational). Then $$\lim\limits_{n \to \infty} x^{q_n r_n} = \lim\limits_{k \to \infty} (\lim\limits_{n \to \infty} x^{q_n r_k}).$$

I don't know how to prove this. I am not even sure that it is true, although I strongly suspect it is. Could anyone help me out here? Any hints would be appreciated. Ideally the proof uses only properties of convergent sequences and rational exponents.

To give some context, I am trying to prove the identity $(x^q)^r = x^{qr}$ for $x,q,r$ real and $x$ positive, using the definition that if $(a_n)_{n=1}^{\infty}$ a sequence of rational numbers with $\lim_{n \to \infty} a_n = a$, and $x > 0$ is a real number, then $$x^a = \lim\limits_{n \to \infty} x^{a_n}.$$

I am attempting to use the following chain of identities, all of which are either true by definition or easily proven, except for the one marked with $?$.

\begin{align}(x^q)^r &= (\lim\limits_{n \to \infty} x^{q_n})^r \\&= \lim\limits_{k \to \infty} (\lim\limits_{n \to \infty} x^{q_n})^{r_k} \\&= \lim\limits_{k \to \infty} (\lim\limits_{n \to \infty} (x^{q_n})^{r_k}) \\&= \lim\limits_{k \to \infty} (\lim\limits_{n \to \infty} x^{q_n r_k}) \\&\stackrel{?}{=} \lim\limits_{n \to \infty} x^{q_n r_n} = x^{qr}.\end{align}

Jonas
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  • Since the logarithm is continuous, you can forget about the $x$, or, if you prefer, use continuity of the function $(y\mapsto x^y)$. Then, notice that both exponents converge to $qr$ where $q$ and $r$ are the limits of $q_n$ and $r_n$. – nietschmann Mar 28 '23 at 13:22
  • It is best to prove that the relation holds for rational $r$ and then $(x^q) ^{r_n} =x^{qr_n} $ for all $n$. Taking limits we get desired identity. The result for rational $r$ is an algebraic consequence of $x^{a+b} =x^ax^b$. – Paramanand Singh Mar 29 '23 at 01:48
  • An essential ingredient needed here is to prove that $x^r\to 1$ when $r\to 0$ and $r$ being a real variable. – Paramanand Singh Mar 29 '23 at 02:24
  • @ParamanandSingh Thanks for your reply! Unfortunately I'm not sure if this works, or I may misunderstand what you're saying. I cannot just take the limit with $x^{q r_n}$: here the exponent is not rational (since $q$ is real), so the definition does not apply. This is the part which makes all this so tricky. – Jonas Mar 29 '23 at 06:56
  • @JonasE.: that's why I wrote the second comment about limit of $x^r$ being $1$. – Paramanand Singh Mar 29 '23 at 09:19
  • Let us write $p_n=(q-q_n) r_n$ so that $p_n\to 0$ and $x^{p_n} \to 1$ and then $x^{qr_n} =x^{p_n} x^{q_nr_n} \to x^{qr} $. – Paramanand Singh Mar 29 '23 at 12:25

1 Answers1

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Yes, the equality is true, since both sequences $(q_n)_n$ and $(r_n)_n$ are convergent. Indeed, fix $x> 0$ and observe that the function $t \in \mathbb{R} \mapsto x^t$ is continuous. Now, let $q = \lim q_n$ and $r= \lim r_n.$ Then, $\lim_n x^{q_nr_n} = x^{\lim_n q_nr_n} = x^{qr}.$

On the other hand, observe that, again by continuity, for each $n \in \mathbb{N} $ it follows that $\lim_k x^{q_nr_k} = x^{q_nr},$ so it is enough to take again the limit $\lim_n x^{q_nr} = x^{qr}$ to obtain the result!

javi1996
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    Observe that the answer is no longer true if you drop the assumption of convergence of any of the sequences: you could take $r_n = n$ and $q_n = 1/n^2$, it is easy to see that the right limit is $\infty$ if $x>1.$ – javi1996 Mar 28 '23 at 13:28
  • Thank you for your answer! Unfortunately, I cannot appeal to continuity, since it has not been defined yet (the proof in question is Lemma 6.7.3 in Tao, Analysis I). Sorry, I should have been more clear about that. Really all that I can use are properties of convergent sequences, real numbers raised to rational exponents, and the definition that if $\lim\limits_{n \to \infty} q_n = q$ and $x > 0$, then $\lim\limits_{n \to \infty} x^{q_n} = x^q$ – Jonas Mar 28 '23 at 13:42
  • The only property of continuity we are using here is the one that you are pointing out, that is, that if $\lim_n q_n =q$ and $x>0$, then $\lim_n x^{q_n} = x^q$! Just recall that the same holds for $\lim x^{q_nr_n} = x^{qr}$ since $\lim_n q_nr_n = qr$ by the elementary properties of limits. – javi1996 Mar 28 '23 at 13:48
  • Fair enough! The one step I am not sure about is $\lim_n x^{q_nr} = x^{qr}$. Here the sequence in the exponent is not rational, so I cannot apply the definition. Could you clarify this? Sorry if I'm being slow here; this kind of reasoning is still new to me. – Jonas Mar 28 '23 at 13:53
  • To clarify, the definition is that if the $q_n$ are rational and $\lim\limits_{n \to \infty} q_n = q $, then $\lim\limits_{n \to \infty} x^{q_n} = x^q$. Again, sorry for being unclear. – Jonas Mar 28 '23 at 14:00
  • Then consider $x^q$ as your new $x$ and use the law of exponents $(x^a)^b = x^{ab}$ which I am sure you have proven before. You have $\lim_n x^{q_nr} = \lim_n (x^r)^{q_n} = (x^r)^q = x^{qr}$. – nietschmann Mar 28 '23 at 14:19
  • @nietschmann Thank you for your comment. Actually this law is exactly what I am trying to prove! That's how the original identity came up in the first place. We have $(x^q)^r = (\lim\limits_{n \to \infty} x^{q_n})^r = \lim\limits_{k \to \infty} (\lim\limits_{n \to \infty} x^{q_n})^{r_k} = \lim\limits_{k \to \infty} (\lim\limits_{n \to \infty} x^{q_n r_k}) \stackrel{?}{=} \lim\limits_{n \to \infty} x^{q_n r_n} = x^{qr}$. – Jonas Mar 28 '23 at 15:16