<s>[INST] 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? 
 [/INST] $$\ln\sqrt{1+x^2}=\dfrac{\ln(1+x^2)}2=\dfrac12\left(x^2-\dfrac{x^4}2+\dfrac{x^6}3-\cdots\right)$$
Now $1+\sin x=\left(\cos\dfrac x2+\sin\dfrac x2\right)^2$
Now $\cos\dfrac x2+\sin\dfrac x2=\sqrt2\sin\left(\dfrac x2+\dfrac \pi4\right)>0$ for $x\to0$
$\implies\sqrt{1+\sin x}=\cos\dfrac x2+\sin\dfrac x2$
and $\cos\dfrac x2=1-\dfrac{(x/2)^2}{2!}+\dfrac{(x/2)^2}{4!}-\cdots$
$\sin\dfrac x2=x/2-\dfrac{(x/2)^3}{3!}+\dfrac{(x/2)^5}{5!}-\cdots$
Dord it help?
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