Proving $\lim\limits_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right) =-\frac12$ How can I prove that 

$$\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)=-\frac{1}{2}$$

 A: $$\begin{aligned}\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)
&=\lim _{x\to 0}\left(\frac{\ln \left(x+1\right)-\ln \left(x+\sqrt{x^2+1}\right)}{\ln \left(x+1\right)\ln \left(x+\sqrt{x^2+1}\right)}\right)
\\&=\lim _{x\to 0}\left(\frac{x-\frac{1}{2}x^2+o\left(x^2\right)-\left(x+o\left(x\right)\right)}{\left(x-\frac{1}{2}x^2+o\left(x^2\right)\right)\left(x+o\left(x\right)\right)}\right)
\\&=\lim _{x\to 0}\left(\frac{-\frac{x^2}{2}+o\left(x\right)}{x^2-\frac{x^3}{2}+o\left(x\right)}\right)
\\&=\color{red}{-\frac{1}{2}}\end{aligned}$$
Solved with Taylor expansions
A: The answer can be achieved using simple limit theorems as follows
$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\left(\frac{1}{\log(x + \sqrt{1 + x^{2}})} - \frac{1}{\log(1 + x)}\right)\\
&= \lim_{x \to 0}\frac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{\log(x + \sqrt{1 + x^{2}})\log(1 + x)}\\
&= \lim_{x \to 0}\frac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{\log(1 - 1 + x + \sqrt{1 + x^{2}})\log(1 + x)}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{{\displaystyle \left(x + \sqrt{1 + x^{2}} - 1\right)\dfrac{\log\left\{1 + \left(x + \sqrt{1 + x^{2}} - 1\right)\right\}}{x + \sqrt{1 + x^{2}} - 1}\cdot x\cdot\dfrac{\log(1 + x)}{x}}}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{{\displaystyle \left(x + \sqrt{1 + x^{2}} - 1\right)1\cdot x\cdot 1}}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) - \log\left(\dfrac{\sqrt{1 + x^{2}} - x}{\sqrt{1 + x^{2}} - x}\cdot\left(x + \sqrt{1 + x^{2}}\right)\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log(1 + x) + \log\left(\sqrt{1 + x^{2}} - x\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log\left\{(1 + x)\left(\sqrt{1 + x^{2}} - x\right)\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log\left\{\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2}\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log\left\{1 - 1 + \sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2}\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\log\left\{1 + \left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)\right\}\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)}{\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{\left(x^{2} + x\sqrt{1 + x^{2}} - x + \sqrt{1 + x^{2}}  - 2x^{2} - 1\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= \lim_{x \to 0} 1 + \dfrac{\sqrt{1 + x^{2}}  - 2x^{2} - 1}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\\
&= 1 + \lim_{x \to 0}\dfrac{1 + x^{2} - (2x^{2} + 1)^{2}}{x \left(1 + x^{2} - (1 - x)^{2}\right)}\frac{\sqrt{1 + x^{2}} + 1 - x}{\sqrt{1 + x^{2}} + 2x^{2} + 1}\\
&= 1 + \lim_{x \to 0}\dfrac{-3x^{2} - 4x^{4}}{2x^{2}}\cdot\frac{2}{2} = 1 - \frac{3}{2} = -\frac{1}{2}\end{aligned}$
The above solution looks long because of detailed steps involving algebraic simplification but involves nothing more than logarithmic limit $\lim\limits_{y \to 0}\dfrac{\log(1 + y)}{y} = 1$ and rationalization to handle $\sqrt{1 + x^{2}}$. 
