Power series of $\frac{\sqrt{1-\cos x}}{\sin x}$ When I'm trying to find the limit of  $\frac{\sqrt{1-\cos x}}{\sin x}$ when x approaches 0, using power series with "epsilon function" notation, it goes :
$\dfrac{\sqrt{1-\cos x}}{\sin x} =  \dfrac{\sqrt{\frac{x^2}{2}+x^2\epsilon_1(x)}}{x+x\epsilon_2(x)}  =   \dfrac{\sqrt{x^2(1+2\epsilon_1(x))}}{\sqrt{2}x(1+\epsilon_2(x))} = \dfrac{|x|}{\sqrt{2}x}\dfrac{\sqrt{1+2\epsilon_1(x)}}{1+\epsilon_2(x)} $
But I can't seem to do it  properly  using Landau notation
I wrote :
$ \dfrac{\sqrt{\frac{x^2}{2}+o(x^2)}}{x+o(x)} $
and I'm stuck... I don't know how to carry these o(x) to the end
Could anyone please show me what the step-by-step solution using Landau notation looks like when written properly ?

Now
$$\frac{\sqrt{1 - \cos(x)}}{\sin(x)} = \frac{\sqrt{2 \sin^2(x/2)}}{\sin(x)}
= \sqrt{2} \frac{ |\sin(x/2)|}{\sin(x)} = \sqrt{2} \frac{|x/2| + O(x^3)}{x + O(x^3)}\\
= \sqrt{2} \frac{\text{sgn}(x)/2 + O(x^2)}{1 + O(x^2)} = \text{sgn}(x)/\sqrt{2} + O(x^2)$$
