Where does this identity follow from Taylor's theorem?

Question

In my complex analysis textbook (Freitag-Busam), we are asked to prove the following theorem:

Theorem: Let $f,g:D\to\mathbb{C}$ be analytic functions, which have the same order $k$ in $a\in D$. Then $h:=f/g$ has in $a$ a removable singularity, and the following formula holds: $$\lim_{z\to a}\frac{f(z)}{g(z)}=\frac{f^{(k)}(a)}{g^{(k)}(a)}$$

We are given the hint to write $f(z)=(z-a)^kf_0(z)$ and then use Taylor's theorem to conclude that $f_0(a)=f^{(k)}(a)/k!$. However, I have been unable to see how to use Taylor's theorem to derive this identity. I attempted to prove it using the Cauchy integral theorem and the fact that $f_0$ is analytic at $a$ as follows, for $\gamma$ suitably chosen around $a$: $$f_0(a)=\frac{1}{2\pi i}\oint_\gamma\frac{f_0(z)}{z-a}dz=\frac{1}{2\pi i}\oint_\gamma\frac{(z-a)^kf_0(z)}{(z-a)^{k+1}}dz=\frac{1}{k!}\big((z-a)^kf_0(z)\big)^{(k)}(a)$$ But since $(z-a)^kf_0(z)=f(z)$, this is equivalent to $\frac{1}{k!}f_0^{(k)}(z)$.

However I am still not sure how to prove the identity using Taylor's theorem. Any ideas? We do not need to use Taylor's theorem, but I am curious how to prove the identity without the Cauchy integral formula.

Answer 1

For simplicity, let us agree that $a = 0$. The formulation of Taylor's Theorem in complex analysis I am used to is that

$$f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n.$$

If $f$ and the first $k-1$ derivatives of $f$ are zero at zero, then it follows that $f(z) = z^k f_0(z)$, as you say. But then, in order for the $k$-th coefficient to match up with the Taylor series above, we must have $f_0(0) = f^{(k)}(0)/k!.$

Answer 2

By Taylor's theorem, $$ f(z) = \sum_{n=k}^\infty\frac{f^{(n)}(a)}{n!}(z-a)^n = (z-a)^k\left(\frac{f^{(k)}(a)}{k!} + (z-a)\sum_{\ell=0}^\infty\frac{f^{(k+1k\ell)}(a)}{(k+1+\ell)!}(z-a)^\ell\right). $$ The series is analytic at $a$ and thus bounded. So, $\lim_{z\to a}\frac{f(z)}{(z-a)^k} = \frac{f^{(k)}(a)}{k!}$.