Prove that $f$ has a removable singularity at $z_0$, and compute $\lim_{z\to z_0} f(z)$ I am again stuck on a qual question while I am preparing for my upcoming exam:
Let $W$ be analytic in a domain $D$. Let $z_0\in D$ be such that $W'(z_0)\neq 0$. Define
$$f(z) = \dfrac{W'(z)W'(z_0)}{(W(z)-W(z_0))^2} - \dfrac{1}{(z-z_0)^2}, z\in D\backslash \{z_0\}$$
Prove that $f$ has a removable singularity at $z_0$, and compute $\displaystyle \lim_{z\to z_0} f(z)$.
My attempt:
I tried rewriting $f(z)$ by using the idea that $W'(z_0) = \displaystyle \lim_{z\to z_0} \dfrac{W(z)-W(z_0)}{z-z_0}$. So I got $f(z) \sim \dfrac{W'(z)-W'(z_0)}{(z-z_0)(W(z)-W(z_0))}$. However, I still can't seem to get the limit.
 A: By Taylor expansion about $z_0$, you can formally compute the limit, but it is far from elegant. 
Write
\begin{align}
f(z) = \frac{W'(z)W'(z_0) - \left[\frac{W(z)-W(z_0)}{(z-z_0)}\right]^2}{[W(z)-W(z_0)]^2}.
\end{align}
The Taylor expansion
\begin{equation}
W(z)-W(z_0) = W'(z_0)(z-z_0) + \frac{1}{2}W''(z_0)(z-z_0)^2 + \ldots
\end{equation}
gives
\begin{equation}
\left[W(z)-W(z_0)\right]^2 = W'(z_0)^2(z-z_0)^2 + W'(z_0)W''(z_0)(z-z_0)^3 + \ldots
\end{equation}
and also
\begin{equation}
\left[\frac{W(z)-W(z_0)}{z-z_0}\right]^2 = W'(z_0)^2 + W'(z_0)W''(z_0)(z-z_0) + \ldots
\end{equation}
where '$\ldots$' indicate higher powers of $(z-z_0)$.
Therefore
\begin{equation}
W'(z)W'(z_0) - \left[\frac{W(z)-W(z_0)}{z-z_0}\right]^2 = W'(z_0)\left[W'(z) - W'(z_0) - W''(z_0)(z-z_0) + \ldots\right]
\end{equation}
The term in brackets in the above can be written
\begin{equation}
W'(z) - W'(z_0) - W''(z_0)(z-z_0) = \frac{1}{2}W'''(z_0)(z-z_0)^2 + \ldots
\end{equation}
With these expansions, we see
\begin{equation}
f(z) = \frac{\frac{1}{2}W'(z_0)W'''(z_0)(z-z_0)^2 + \ldots}{W'(z_0)^2(z-z_0)^2 + \ldots}
\end{equation}
Thus $\lim_{z\rightarrow z_0}f(z) = \frac{W'''(z_0)}{2W'(z_0)}$. 
