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.

i will use the maclaurin expansion for $\sqrt{1+x}, \ln(1+x), \tan(x)$
$\begin{align}
\ln[x + (1 + x^2)^{1/2}] &= \ln[x + 1 + \frac{1}{2}x^2 -\frac{1}{8}x^4 + \cdots]\\
&=\ln(1 + x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots)  \\
&=(x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots)-\frac{1}{2} \{x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots\}^2 \\
&+ \frac{1}{3}\{x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots\}^3  \cdots\\
&=(x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots)-\frac{x^2}{2} (1 + x + \cdots) +  \frac{1}{3}(x^3  + \cdots) +\cdots\\
&=x-\frac{1}{2}x^3 + \frac{1}{3}x^3 + \cdots\\
&= x-\frac{1}{6}x^3 + \cdots
\end{align}$
the expansion for 
$$\tan x = x + \cdots$$  putting the two together 
$$\lim_{x \to 0}\dfrac{\ln[x + (1 + x^2)^{1/2}] - x}{\tan^3 x} = -\dfrac{1}{6}$$
