Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ Evaluate the given limit:
$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$
I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show me what is the trick here?
Thanks.

This solution is based on no l'Hospital rule nor Taylor expansion. The following standard limits only are used:
\begin{eqnarray*}
\lim_{x\rightarrow 0}e^{x} &=&1. \\
\lim_{x\rightarrow 0}\frac{\tan x}{x} &=&1. \\
\lim_{x\rightarrow 0}\frac{1+x+\frac{1}{2}x^{2}-e^{x}}{x^{3}} &=&-\frac{1}{6}%
. \\
\lim_{x\rightarrow 0}\frac{\sqrt{1+x^{2}}-1-\frac{1}{2}x^{2}}{x^{3}} &=&0.
\end{eqnarray*}
We transform the original expression $f(x)=\dfrac{\ln (x+\sqrt{1+x^{2}})-x}{\tan
^{3}x}$ as follows
\begin{eqnarray*}
f(x)&=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (x+\sqrt{1+x^{2}})-\ln e^{x}}{%
x^{3}} \\
&=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (\dfrac{x+\sqrt{1+x^{2}}}{e^{x}})}{%
x^{3}} \\
&=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (1+\color{red}{u(x)})}{\color{red}{u(x)}}\cdot \dfrac{\color{red}{\dfrac{%
x+\sqrt{1+x^{2}}-e^{x}}{e^{x}}}}{x^{3}},\ \ \ with\ \color{red}{u(x)=\frac{x+\sqrt{1+x^{2}}-e^{x}}{e^{x}}}  \\
&=&\left( \frac{x}{\tan x}\right) ^{3}\cdot \dfrac{\ln (1+u(x))}{u(x)}\cdot 
\dfrac{1}{e^{x}}\cdot \left( \left( \frac{\sqrt{1+x^{2}}\color{green}{-1-\frac{1}{2}x^{2}}}{%
x^{3}}\right) +\left( \frac{\color{blue}{1}+x\color{blue}{+\frac{1}{2}x^{2}}-e^{x}}{x^{3}}\right)
\right) 
\end{eqnarray*}
Since $\lim\limits_{x\rightarrow 0}u(x)=0$,  and the function $t\rightarrow t^3$ is continuous, then
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\ln (x+\sqrt{1+x^{2}})-x}{\tan ^{3}x} &=&\left(
\lim_{x\rightarrow 0}\frac{x}{\tan x}\right) ^{3}\cdot \lim_{u\rightarrow 0}%
\frac{\ln (1+ u)}{u}\cdot \lim_{x\rightarrow 0}\frac{1}{e^{x}} \\
&&\cdot \left( \left( \lim_{x\rightarrow 0}\frac{\sqrt{1+x^{2}}-1-\frac{1}{2}%
x^{2}}{x^{3}}\right) +\left( \lim_{x\rightarrow 0}\frac{1+x+\frac{1}{2}%
x^{2}-e^{x}}{x^{3}}\right) \right)  \\
&=&\left( 1\right) ^{3}\cdot 1\cdot \frac{1}{1}\cdot \left( \left( 0\right)
+\left( -\frac{1}{6}\right) \right)  \\
&=&\color{red}{-\frac{1}{6}}.
\end{eqnarray*}
