McLaurin expansions help

Question

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(!)

Answer 1

Make them into one fraction.

$\frac {\ln (1+ 2x) - \ln (2x + \sqrt{1+x^3})}{\ln (1+ 2x) \ln (2x + \sqrt{1+x^3})}$

Expand the series for $\sqrt {1+x^3} = 1 + \frac 12 x^3 - \frac 18 x^6 \cdots$

$\frac {\ln (1+ 2x) - \ln ( 1 + 2x + \frac 12x^3\cdots)}{\ln (1+ 2x) \ln (1+2x + \frac 12 x^3 \cdots )}$

and then the logs.

$\ln(1+2x) = 2x - \frac 12 (2x)^2 + \frac 13 (2x)^3 \cdots\\ \ln(1+2x+\frac 12 x^3) = 2x + \frac 12 x^3 - \frac 12 (2x + \frac 12x^3)^2 \cdots $

the difference $\frac 12 x^3 - x^4\cdots$

The product $4x^2 - 8x^3 \cdots$

So it would appear that the numerator is of order $x^{a+3}$ and the denominator is of order $x^2$

The limit coverges to 0 if $x > -1$ and diverges if $a < -1$ and converges to $-\frac 18$ if $a = -1$

Answer 2

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.