Why Taylor expansion works when doing it in parts and steps, and how to know how many terms to keep?

Question

I am majoring in Physics, and we use Taylor expansions and approximations all the time, so I wanted to understand how they work better. For example, in one of my problems, I was dealing with the following function: $$ f(x) = \frac{\sin{x}}{\sqrt{2 - \sqrt{2 + 2\cos{x}}}} $$ for $x > 0$ and more importantly, near zero: $x \ll 1$. The original problem was to find the limit of this function as $x \to 0^+$. I tried using L'Hopital's rule and some other limit tricks, but it quickly became a mess you can't get out of. Then, I looked at the answer, and I saw that they solved the limit using the famous Taylor approximations: $$ \sin{x} \approx x \qquad \cos{x} \approx 1 - \frac{x^2}{2} \qquad (1 + x)^{\alpha} \approx 1 + \alpha x $$ Plugging these approximations in, we get: \begin{align} \lim_{x \to 0^+}\frac{\sin{x}}{\sqrt{2 - \sqrt{2 + 2\cos{x}}}} &\approx \lim_{x \to 0^+}\frac{x}{\sqrt{2 - \sqrt{2 + 2(1 - \frac{x^2}{2})}}} \\ &= \lim_{x \to 0^+}\frac{x}{\sqrt{2 - \sqrt{4 - x^2}}} \\ &= \lim_{x \to 0^+}\frac{x}{\sqrt{2 - 2\sqrt{1 - \frac{x^2}{4}}}} \\ &\approx \lim_{x \to 0^+}\frac{x}{\sqrt{2 - 2(1 - \frac{x^2}{8})}} \\ &= \lim_{x \to 0^+}\frac{x}{\sqrt{\frac{x^2}{4}}} \\ &= 2 \end{align} This is indeed the correct result. Honestly, this approach looks like magic to me, and it looks like it can solve many tough limit problems easily. We basically found the zeroth order Taylor approximation of $f(x)$. This got me thinking that if I keep more terms in my approximations, then I should be able to produce the rest of the Taylor series of $f(x)$. So I tried keeping the next terms of the trig functions: $$ \sin{x} \approx x - \frac{x^3}{6} \qquad \cos{x} \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} $$ Repeating the same calculation, I get: \begin{align} \frac{\sin{x}}{\sqrt{2 - \sqrt{2 + 2\cos{x}}}} &\approx \frac{x - \frac{x^3}{6}}{\sqrt{2 - \sqrt{2 + 2(1 - \frac{x^2}{2} + \frac{x^4}{24})}}} \\ &= \frac{x - \frac{x^3}{6}}{\sqrt{2 - \sqrt{4 - x^2 + \frac{x^4}{12}}}} \\ &= \frac{x - \frac{x^3}{6}}{\sqrt{2 - 2\sqrt{1 - \frac{x^2}{4} + \frac{x^4}{48}}}} \\ &\approx \frac{x - \frac{x^3}{6}}{\sqrt{2 - 2(1 - \frac{x^2}{8} + \frac{x^4}{96})}} \\ &= \frac{x - \frac{x^3}{6}}{\sqrt{\frac{x^2}{4} - \frac{x^4}{48}}} \\ &= \frac{2 - \frac{x^2}{3}}{\sqrt{1 -\frac{x^2}{12}}} \\ &\approx \left(2 - \frac{x^2}{3}\right)\left(1 + \frac{x^2}{24}\right) \\ &\approx 2 - \frac{x^2}{4} \end{align} While I got that the linear term correctly vanishes, I got the wrong coefficient for the quadratic term. That's when I realized I should also keep another term from the third approximation: $$ (1 + x)^{\alpha} \approx 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2}x^2 $$ Repeating the calculation again with this improved approximation: \begin{align} \frac{\sin{x}}{\sqrt{2 - \sqrt{2 + 2\cos{x}}}} &\approx \frac{x - \frac{x^3}{6}}{\sqrt{2 - 2\sqrt{1 - \frac{x^2}{4} + \frac{x^4}{48}}}} \\ &\approx \frac{x - \frac{x^3}{6}}{\sqrt{2 - 2(1 + \frac{1}{2}(-\frac{x^2}{4}+ \frac{x^4}{48}) - \frac{1}{8}(-\frac{x^2}{4}+ \frac{x^4}{48})^2)}} \\ &\approx \frac{x - \frac{x^3}{6}}{\sqrt{\frac{x^2}{4} - \frac{x^4}{192}}} \\ &= \frac{2 - \frac{x^2}{3}}{\sqrt{1 - \frac{x^2}{48}}} \end{align} Now, I noticed that to get a second-order approximation, it's enough to take the linear approximation of the square root in the denominator (keeping the quadratic term will only give me higher-order terms): \begin{align} \frac{\sin{x}}{\sqrt{2 - \sqrt{2 + 2\cos{x}}}} &\approx \frac{2 - \frac{x^2}{3}}{\sqrt{1 - \frac{x^2}{48}}} \\ &\approx \left(2 - \frac{x^2}{3}\right)\left(1 + \frac{x^2}{96}\right) \\ &\approx 2 - \frac{5x^2}{16} \end{align} This is indeed the correct second-order Taylor expansion of my function $f(x)$. Now I have a few questions about this whole process:

1. As a physicist, I am used to calculating Taylor approximations like this and it always felt intuitive to me. However, now I suddenly realize it's not as trivial as I thought. My first and most important question is why does this even work in the first place?? The Taylor expansion is given by: $$ f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2}x^2 $$ We only used this formula for the basic expansions of $\sin{x}$, $\cos{x}$ and $(1 + x)^{\alpha}$, but we didn't compute $f$ or any of its derivatives at $x = 0$. It certainly looks like a nightmare to differentiate $f(x)$ even once, let alone twice. Besides, I still have to compute limits to get $f$ and its derivatives at $x = 0$, which I couldn't manage to do without the Taylor expansions to begin with. Why can I take a function and expand parts of it and then continue expanding by parts and making small steps until I get to the final result instead of going through Taylor's formula? The fact it just works honestly looks like magic to me.
2. How do I know how many terms I need to keep for each part when expanding? In the beginning, I wanted to get a zeroth-order approximation of $f(x)$, but I had to use the first-order approximations of $\sin{x}$ and $(1 + x)^{\alpha}$ and the second-order approximation for $\cos{x}$. Later, when I tried to get the second-order approximation for $f(x)$, I had to use the second-order approximation of $\sqrt{1 + x}$, the 3rd order approximation of $\sin{x}$ and the 4th order approximation of $\cos{x}$. In the final step of the calculation, I only needed the first-order approximation of $\frac{1}{\sqrt{1 + x}}$ and not the second order. Is there a way to know how many terms I need to keep for each approximation? As you saw in the calculation, when I took too few terms of some approximations, I got a wrong answer, but I couldn't know it was wrong (As opposed to the zeroth order. If I took too few terms there, I'd quickly get zero in the numerator or denominator, signaling that something is wrong).
3. Is there a meaning to the second calculation I made when I got $f(x) \approx 2 - \frac{x^2}{4}$? From my point of view, I still took correct approximations, so I should still get a good approximation for $f(x)$. It's not the Taylor approximation, but is this still good for something, or did I just get a garbage result?
4. Is there a more formal or precise way to make these calculations to avoid wrong results like I got? As mentioned before, I am a physicist, but I am willing to learn and apply math properly, especially if it helps me make correct calculations and gives me insight or intuition for what I do. I am curious to know how a mathematician would make these calculations and if it could help me make similar calculations more easily and correctly.

Answer 1

$$f(x) = \frac{\sin(x)}{\sqrt{2 - \sqrt{2 + 2\cos(x)}}}$$ Lok at the denominator and use (from inside) $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^ 6}{720}+O\left(x^8\right)$$ $$2+2\cos(x)=4-x^2+\frac{x^4}{12}-\frac{x^6}{360}+O \left(x^8\right)$$ $$\sqrt{2 + 2\cos(x)}=2-\frac{x^2}{4}+\frac{x^4}{192}-\frac{x ^6}{23040}+O\left(x^8\right)$$ $$2-\sqrt{2 + 2\cos(x)}=\frac{x^2}{4}-\frac{x^4}{192}+\frac{x^6}{23040}+O\left(x^8\right)$$

Assuming $x>0$ $$\sqrt{2 - \sqrt{2 + 2\cos(x)}}=\frac{x}{2}-\frac{x^3}{192}+\frac{x^5}{61440}+O\left(x^6\right)$$ Now, use the series expansion for the numerator to have $$f(x)=\frac {x-\frac{x^3}{6}+\frac{x^5}{120}+O\left( x^7\right) } {\frac{x}{2}-\frac{x^3}{192}+\frac{x^5}{61440}+O\left(x^6\right) }$$ Now, the long division to get $$f(x)=2-\frac{5 x^2}{16}+\frac{41x^4}{3072}+O\left(x^5\right)$$ which gives the limit and much more.

Suppose that you want to solve $f(x)=1$; the above truncated series gives $$x^2=\frac{32}{41}\left(15-\sqrt{102}\right)$$ that is to say $x=1.95570$ whle the solution is $1.94025$. This is a very good estimate for starting any root finding method.

Edit

For applications in physics, let $S_n$ be the expansion to $O(x^{2n+2})$ and suppose that we want to know $n$ for covering the range $0\leq x \leq \pi$.

Consider the infinite norm $$\Phi_n=\int_0^\pi \Big(f(x)-S_n \Big)^2\,dx$$ to be computed numerically

$$\left( \begin{array}{cc} n & \Phi_n \\ 1 & 4.3885\times 10^{-1} \\ 2 & 1.1552\times 10^{-2} \\ 3 & 9.1993\times 10^{-5} \\ 4 & 2.9281\times 10^{-7} \\ 5 & 4.4387\times 10^{-10} \\ 6 & 3.6072\times 10^{-13} \\ 7 & 1.7115\times 10^{-16} \\ \end{array} \right)$$

Bsed on that, make your decision; mine would be $n=5$.