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.

L'Hospital's rule:
$$\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{\tan^3(x)}=\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{x^3}\cdot\frac{\tan^3 x}{x^3}=$$
$$=\lim_{x\to0} \frac{(\ln(x+\sqrt{1+x^2})-x)'}{(x^3)'}=\lim_{x\to0}\frac{\frac{1}{\sqrt{1+x^2}}-1}{3x^2}=\lim_{x\to0}\frac{-x^2}{3x^2\sqrt{1+x^2}(1+\sqrt{1+x^2})}=-\frac{1}{6}.$$
