Solution of $\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $ using Mclaurin series The problem is 
$$\lim_{x\to 0} \frac{(x  \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $$
I know that 
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}... $$ 
got this 
$$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15}...$$ 
and suppose to use taylor expansion for  $\ln {(1 + x)}$
Tried to expand $\sqrt{1 + \sin x}$ and $\ln{\sqrt{(1 + x^2)}}$  but it seems to be  messy. Am I on the right direction? 

Note that $$x\sqrt{1+\sin x} = x\sqrt{1+x+O(x^3)} = x(1+\frac x2-\frac{x^2}8+O(x^3)) = x+\frac{x^2}{2}-\frac{x^3}8+O(x^4)$$
and
$$\ln\sqrt{1+x^2} = \ln(1+\frac{x^2}2+O(x^4)) = \frac{x^2}{2}+O(x^4)$$
so we have
$$\lim_{x\to 0}\frac{x\sqrt{1+\sin x}-\ln\sqrt{1+x^2}-x}{\tan^3x} = \lim_{x\to 0}\frac{-\frac{x^3}8+O(x^4)}{x^3+O(x^4)} = -\frac 18$$
