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?
