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?
On
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
$$\lim_{x\to 0}\frac{x\sqrt{1+\sin x}-\ln\sqrt{1+x^2}-x}{\tan^3x} = \lim_{x\to 0}\frac{-\frac{x^3}8+O(x^4)}{x^3+O(x^4)} = -\frac 18$$
$$\ln\sqrt{1+x^2}=\dfrac{\ln(1+x^2)}2=\dfrac12\left(x^2-\dfrac{x^4}2+\dfrac{x^6}3-\cdots\right)$$
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?