Having trouble understanding why the $r$-th mean tends to the geometric mean as $r$ tends to zero

Question

I am having trouble understanding the proof of Theorem 3 in "Inequalities" by Hardy, Littlewood and Pólya. This theorem states that the $r$-th mean approaches the geometric mean as $r$ approaches zero.

I have seen the following post which makes things a little clearer (albeit using $o(r)$ instead of $O(r^2)$):

Why is the $0$th power mean defined to be the geometric mean?

However, I still cannot determine why:

(a): $a^r = 1 + r\log(a) + O(r^2)$ as $r$ tends to zero,

and

(b): $\lim_{r\to 0} (1 + rx + o(r))^{1/r} = e^x$.

I have a pretty solid grasp of limits, as well as the log and exp functions, but I have never really been taught anything substantial on big/little-O notation, in particular as the variable approaches zero. Could somebody point me towards a suitable proof of (a) and (b) above please.

Answer 1

a) Use the definition of $a^r$ and the well known Taylor series for $e^x$ $$a^r = (e^{\log a})^r = e^{r\log a} = 1 + r\log a + r^2(\log a)^2 \dots $$ b) Hint: Do you remember this formula? $$\lim_{n\to \infty} (1 + \frac{x}{n})^n = e^x$$

Answer 2

a) $\lim_{r\to 0}\frac {α^r-1-rlogα}{r^2}=\lim_{r\to 0} \frac {α^rlogα-logα}{2r}=\lim_{r\to 0} \frac {α^r log^2α}{2}=\frac {log^2α}{2} \implies α^r-1-rlogα=O(r^2)$, where we used d. l'H. rule twice for $\frac 00$ forms and the proposition $f(x)=O(g(x))$, as $x\to x_0$ $\Leftarrow\lim_{x\to x_0}\frac fg(x)= c\in \Bbb R.$

b) $\lim_{r\to 0}(1+rx+o(r))^{1/r}=\lim_{r\to 0} (1+\frac {x}{\frac 1r}+o(r))^\frac 1r=\lim_{s\to \infty}(1+\frac {x}{s})^s=e^x$, from the definition of exponential and the proposition $o(r)\to 0$, as $r\to 0$.