<s>[INST] 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.
 [/INST] 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*}
</s>