<s>[INST] McLaurin expansions help The problem:
Study the convergence/divergence of this limit, as $a \in \mathbb{R}$ changes
$$\lim_{x\to 0^+} x^a \biggl( \frac{1}{\log(2x+\sqrt{1+x^3})} -\frac{1}{\log(2x+1)} \biggr)$$
I did that:
$\sqrt{1+x^3}=1+\frac{x^3}{2} + o(x^3)$ so
$\log\left(1 +2x +\frac{x^3}{2} + o(x^3) \right) = 2x - \frac{x^2}{2} +\frac{x^3}{2} +o(x^3)$
and $\log\left(1 +2x \right) = 2x - \frac{x^2}{2} +\frac{8x^3}{3} +o(x^3)$ .
Now how can I expand $\frac{1}{ 2x - \frac{x^2}{2} +\frac{x^3}{2} +o(x^3)}$ and $\frac{1}{ 2x - \frac{x^2}{2} +\frac{8x^3}{3} +o(x^3)}$ ?
I don't find on internet some expansion for these fractions(!)
 [/INST] You can handle the reciprocals like $1/(2x-2x^2+8x^3/3+o(x^3))$ using binomial theorem as $$(1/2x)(1-x+4x^2/3+o(x^2))^{-1}=\frac{1}{2x}\left(1+x-x^2/3+o(x^2)\right)$$ which is the same as $(1/2x)+(1/2)-x/6+o(x)$. Similarly the other expression is $(1/2x)+(1/2)-7x/24+o(x) $ and thus the given expression is equal to $-x^{a+1}/8+o(x^{a+1})$ and you can now conclude. 
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