Having trouble understanding why the r-th mean tends to the geometric mean as r tends to zero 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 0th power mean defined to be the geometric mean? However, I still cannot determine why: (a): a^r=1+r.log(a)+o(r) as r tends to zero and (b): lim_(r->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 some

$\frac{20b}{{\left(4b^3\right)}^3}$

Which operation could we perform in order to find the number of milliseconds in a year??
${\displaystyle 60\cdot 60\cdot 24\cdot 7\cdot 365}$
${\displaystyle 1000\cdot 60\cdot 60\cdot 24\cdot 365}$
${\displaystyle 24\cdot 60\cdot 100\cdot 7\cdot 52}$
${\displaystyle 1000\cdot 60\cdot 24\cdot 7\cdot 52?}$

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A) No
B) 0
C) Yes
D) 6

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149600000000 is equal to
A)$1.496\times <!--\; \times \; -->{10}^{11}$
B)$1.496\times <!--\; \times \; -->{10}^{10}$
C)$1.496\times <!--\; \times \; -->{10}^{12}$
D)$1.496\times <!--\; \times \; -->{10}^8$

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simplify ${\displaystyle \sqrt{125n}}$

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