I have a function $f(x)$ for which I am seeking the derivatives at zero: $f^{(n)}(0)$ for each positive integer $n$.
I can express $f(x)$ as an infinite series $$ f(x) = \frac{ a_1(x) }{1!} x + \frac{ a_2(x) }{2!} x^2 + \frac{ a_3(x) }{3!} x^3 + \dots $$ where $ \displaystyle \lim_{x \rightarrow 0} a_n(x) = c_n $ is finite for each positive integer $n$.
The Taylor series (Maclaurin series) expansion for $f(x)$ around $x = 0$ is $$ f(x) = f(0) + \frac{ f'\!(x) }{1!} x + \frac{ f''\!(x) }{2!} x^2 + \frac{ f'''\!(x) }{3!} x^3 + \dots $$
Can I conclude that $ f^{(n)}(0) = c_n $ for each $n$?
Or under what circumstances might I "equate the coefficients in the limit"?
(For the particular function $f(x)$ that I am interested in, I have manually confirmed that $f'\!(0) = c_1$, $f''\!(0) = c_2$, and $f'''\!(0) = c_3$. However the approach does not scale easily to calculating $f^{(n)}(0)$.)
Many thanks in advance.
Edit. I have a "handwavium" proof that I would appreciate comments on.
Lemma. Let $f(x) = u(x)v(x)$ for functions that are $m$-times differentiable on a neighbourhood $S$. Then $$ f^{(m)}(t) = \sum_{k=0}^{m} ( \begin{smallmatrix} m \\ k \end{smallmatrix} ) u^{(k)}(t) v^{(m-k)}(t) $$ on $S$.
Proof of Lemma. By induction on $m$, applying the product rule for differentiation.
Proof of proposition that $ f^{(n)}(0) = c_n $ for each $n$. Applying the lemma to the first series yields $$ f^{(n)}(x) = a_n(x) + x\cdot(\text{stuff that is finite}) $$ so $$ \begin{align*} \lim_{x \to 0} f^{(n)}(x) &= \lim_{x \to 0} \Bigl( a_n(x) + ( x\cdot(\text{stuff that is finite}) \Bigr) \\ &= \lim_{x \to 0} a_n(x) + \lim_{x \to 0} x\cdot(\text{stuff that is finite}) & \text{if both limits are finite} \end{align*} $$ and hence $$ f^{(n)}(0) = c_n $$ by continuity (for the left hand side) and assumption about $a_n(x)$ (for the right hand side).
I think this is a counter example.
Let $ a_1(x)=x+1$, and $ a_n(x)=1 $ for all $ n \geqslant 2 $. Your conditions are met. In this case $ f(x)=e^x+x^2−1 $. Then $ f(0)=0 $ . $ f'(0)=1=c_1$ but $ f''(0)= 3 ≠c_2$.
To avoid this you could stipulate each $a_n$ is differentiable in the neighbourhood of $0$ and its derivatives of order $n$ and higher are also zero at $x=0$.