Evaluating a trigonometric limit What is the limit as $x$ approaches $0$ of: $$\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}?$$ 
We cannot use L'Hôpital's rule or anything advanced like Taylor series. I reduced it to, by considering the numerator's conjugate:
$$\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{1+x-\cos x}$$ 
But I cannot go further. Please help. 
EDIT: Thinking carefully, I think I can simplify further:
$$\frac{1}{2}\lim_{x \to 0}\frac{\frac{\tan x - \sin x}{x}}{\frac{1-\cos x}{x}+1}=\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{x}$$
And perhaps the solution:
$$\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{x}=\frac{1}{2}(\lim_{x \to 0}\frac{\tan x}{x} - 1)=\frac{1}{2}(\lim_{x \to 0}\frac{\sin x}{\cos x \cdot x}) - 1=\frac{1}{2}(1- 1)=0$$
Is this correct?

$$\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}\sim_0\frac{\sqrt{1+x+\frac{x^3}3} - \sqrt{1+x-\frac{x^3}6}}{x+\frac{x^2}2}\sim_0\frac{\frac{1}{2}\left(\frac{x^3}3+\frac{x^3}6\right)}{x}\sim_0\frac14 x^2$$
so the desired limit is $0$.
Edit Without Taylor series the calculus is tedious: let
$$f(x)=\sqrt{1+\tan x} - \sqrt{1+\sin x}$$
and
$$g(x)=1+x-\cos x$$
then
$$\lim_{x\to0}\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}\frac{x-0}{g(x)-g(0)}=\frac{f'(0)}{g'(0)}$$
and with painful calculus we find that $f'(0)=0$ and easily $g'(0)=1$ so we conclude.
