Question about the 'rigour' of this Taylor Series 'proof' I've been going over some of my old maths notes I made during my A Levels (some 5 years ago)  recently, and came across a 'proof' of the Taylor series taught to me back then, which made perfect, logical sense, given what I knew at the time. I can see in hindsight that it seems a little handwavy; but I'm curious as to how 'rigourous' it would be considered. Could it be considered a 'valid' proof?
I'm open to more complicated ideas now, so I don't mind how in depth people go!
The "proof"

Imagine we have a function $f(x)$ that is assumed (infinitely) differentiable, and has no singularities. The question is- can we have some (approximate) polynomial representation of $f(x)$ around some point $x=a$?
I.e. can we have
$$f(x) \stackrel{?}{=}a_0+a_1(x-a)+a_2(x-a)^2+a_3(x-a)^3+a_4(x-a)^4+\dots+a_n(x-a)^n+\dots$$
We can deduce the value of each of the constants by finding the zeros:
$$f(a) = a_0$$
But what about the others? We assumed that we can infinitely differentiate $f(x)$, so we have
$$f'(x) = (1)a_1 + 2a_2(x-a) + 3a_3(x-a)^2 + 4a_4(x-a)^3 + \dots + na_n(x-a)^{n-1}+\dots$$
$$f''(x) = 2(1)a_2 + 3(2)a_3(x-a) + 4(3)(x-a)^3 + \dots + n(n-1)a_n(x-a)^{n-2}+\dots$$
$$f'''(x) = 3(2)(1)a_3 + 4(3)(2)(x-a)^2 + \dots + n(n-1)(n-2)a_n(x-a)^{n-3}+\dots$$
and so on.
Recognising the factorial pattern; we can then find:
$$f'(a) = a_1$$
$$f''(a) = 2!a_2$$
$$f'''(a) = 3!a_3$$
$$\dots$$
$$f^{(n)}(a) = n!a_n$$
so that we finally have
$$f(x) = f(a) + f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\dots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\dots$$
And so we can write, with $\Sigma$ notation:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$

Further to this, I ask because when I considered using Taylor series approximations for certain limits, such as $\left(\text{e.g.}\ \ \lim_{x\rightarrow0} \frac{\sin x}{x}\right)$ I ran into some issues.
I know its use here is slightly redundant, because there are many elagant proofs of this limit; but I'm curious since I seemed to get a somewhat circular argument; the 'proof' using Taylor Series revolves around knowing the derivative of $\sin x$, which itself requires the limit.
I take it that maybe Taylor Series is not applicable to a limit like this, and you would need to use other arguments to prove it instead?
I'm curious around other derivations of Taylor Series that would avoid this.
Many thanks!
 A: This is roughly the intuitive idea behind why the Taylor series is "supposed" to exist. However, as written it's not a correct proof, because you've proven a statement that is so strong that it's false: it's not true that the Taylor series of a smooth function converges to it, as Greg Martin says in the comments. There are two different issues here: the Taylor series may not converge at all (which happens if you go outside the radius of convergence; consider e.g. the Taylor series $\frac{1}{1 - x} = \sum_{n \ge 0} x^n$ evaluated at $x = 2$), and it may converge but not to the function (which happens for $f(x) = e^{-\frac{1}{x^2}}$, basically because it decays so quickly at $x = 0$ that all of its Taylor coefficients there are zero).
The rigorous way to proceed is to prove some form of Taylor's theorem with remainder, which measures how accurately the finite Taylor polynomials $\sum_{i=0}^n \frac{f^{(i)}(a)}{i!} (x - a)^i$ approximate $f(x)$ as a function of $n$, then investigate conditions under which the accuracy of the approximation converges to $0$ as $n \to \infty$.
This doesn't really have anything to do with your specific question about $\sin x$ (which has a Taylor series which converges to it everywhere). Your specific question is about the circularity of using l'Hopital's rule to compute $\lim_{x \to 0} \frac{\sin x}{x}$. You are correct that using l'Hopital's rule here is circular, in that this limit is by definition $\sin'(0)$, so you have to already know how to evaluate this limit somehow in order to differentiate $\sin x$. To get a non-circular argument depends on how you define $\sin x$. If you prefer a more geometric definition there are geometric arguments, e.g. the squeeze theorem arguments. Otherwise you can directly define $\sin x$ as a power series, or define it in terms of differential equations.
A definition that allows for clean and easy proofs is the following: we define $\cos t$ and $\sin t$ simultaneously as the unique pair of continuously differentiable functions $(c(t), s(t))$ satisfying

*

*$c(0) = 1, s(0) = 0, c'(0) = 0, s'(0) = 1$

*$c(t)^2 + s(t)^2 = 1$

*$c'(t)^2 + s'(t)^2 = 1$.

These conditions together say that the function $t \mapsto (c(t), s(t))$ is a unit speed parameterization of the unit circle $x^2 + y^2 = 1$ starting at $(1, 0)$ and moving counterclockwise. It should be intuitively obvious that this gives both existence and uniqueness; to prove it formally, differentiating the second condition gives that the displacement vector $(c(t), s(t))$ is orthogonal to the velocity vector $(c'(t), s'(t))$, and since they are both unit vectors this means they're always a fixed angle apart, which must be either $\frac{\pi}{2}$ or $\frac{3 \pi}{2}$. The initial conditions fix this angle: $(c'(t), s'(t))$ is $\frac{\pi}{2}$ counterclockwise from $(c(t), s(t))$, which can be written using complex numbers as
$$\text{cis}'(t) = i \text{ cis}(t)$$
where $\text{cis}(t) = \cos t + i \sin t$. Conversely you can show that this condition, together with the initial conditions, implies both the 2nd and 3rd conditions above. So the conditions above are equivalent to the conditions

*

*$\text{cis}(0) = 1$

*$\text{cis}'(t) = i \text{ cis}(t)$
and then existence and uniqueness follow abstractly from the Picard-Lindelof theorem and concretely from the construction and proof of basic properties of the complex exponential (or the matrix exponential). From here you can easily deduce Euler's formula $e^{it} = \text{cis}(t)$ (it falls out of general facts about the matrix exponential), the angle addition formulas, the derivatives of sine and cosine, etc.
