Power series of $\frac{\sqrt{1-\cos x}}{\sin x}$

Question

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 ?

Answer 1

It is the same as in the "$\epsilon$" notation. For numerator, we want $\sqrt{x^2\left(\frac{1}{2}+o(1)\right)}$, which is $|x|\sqrt{\frac{1}{2}+o(1)}$. In the denominator, we have $x(1+o(1))$.

Remark: Note that the limit as $x\to 0$ does not exist, though the limit as $x$ approaches $0$ from the left does, and the limit as $x$ approaches $0$ from the right does.

Answer 2

If you are willing to skip Landau notation, you can just note (when $0<x<\pi$), $\sin x = \sqrt{\sin^2 x}=\sqrt{1-\cos^2 x}$. Then:

$$\frac{\sqrt{1-\cos x}}{\sin x} = \sqrt{\frac{1-\cos x}{1-\cos^2 x}}=\frac{1}{\sqrt{1+\cos x}}$$

Note this is why there is no general limit as $x\to 0$. When $-\pi<x<0$, the value of the expression is $\frac{-1}{\sqrt{1+\cos x}}$. You can only get a limit as $x\to 0+$ or as $x\to 0-$, not a general limit.

Answer 3

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