If $q, r \in \mathbb{R}, x \in \mathbb{R}^+$ then $(x^q)^r=x^{qr}$ I'm stuck on this exercise from Tao's Analysis 1 textbook:
show that if $q, r \in \mathbb{R}, x \in \mathbb{R}^+$ then $(x^q)^r=x^{qr}$.
DEF. (Exponentiation to a real exponent):
Let $x>0$ be real, and let $\alpha$ be a real number.
We define the quantity $x^\alpha$ by the formula $x^\alpha=\lim_{n\to\infty} x^{q_n}$, where $(q_n)_{n=1}^\infty$ is any sequence of rational numbers converging to $\alpha$.
I've already proved any property of real exponentiation when the exponent is a rational number (for example that the property in question holds when $x \in \mathbb{R}^+$ and $q,r \in \mathbb{Q}$) and that $x^q$ (with $x \in \mathbb{R}^+$ and $ q \in \mathbb{R}$) is a positive real number.
What puzzles me is how to get around the fact that we are considering two limits simultaneously, in fact from the definition above it follows that $(x^q)^r=\lim_{n\to\infty}(\lim_{m\to\infty}x^{q_n})^{r_n}$.
(This question: $(x^r)^s=x^{rs}$ for the real case talks about this exercise, but I don't understand how the author can say that $(x^q)^r=\lim_{n\to\infty}(\lim_{m\to\infty}x^{q_n})^{r_n}=\lim_{n\to\infty}\lim_{m\to\infty}((x^{q_n})^{r_n})=\lim_{n\to\infty}\lim_{m\to\infty} x^{q_nr_n}$.)
So, I would appreciate any hints about how to start/carry out its proof.
Best regards,
lorenzo

$ \newcommand{\seqlim}[1]{\lim\limits_{#1\to\infty}} $
I find a simple answer for this old question.
Show that $ (x^s)^r = x^{sr} $ for $ s\in\mathbb{Q},r\in\mathbb{R} $
$$(x^s)^r=\seqlim{n}{(x^s)^{r_n}}=\seqlim{n}{x^{sr_n}}=a^{sr},\quad [\seqlim{n}sr_n=sr] $$
Then we have
$$ (x^q)^r=\seqlim{n}(x^q)^{r_n}=\seqlim{n}x^{qr_n}=x^{qr},\quad [\seqlim{n}qr_n=qr] $$
