L'Hospital's Rule but the process does not stop Let $\Psi:\mathbb{C}\rightarrow \mathbb{C}$ be an infinitely differentiable function at $\Psi^{-1}(a).$
What is the limit:
$$\lim_{w\rightarrow \Psi^{-1}(a)} \frac{1}{(\Psi(w)-a)(\Psi')^{1/2}(w)}+\frac{1}{(\Psi'(\Psi^{-1}(a)))^{3/2}(\Psi^{-1}(a)-w)}?$$
I kept using L Hôpital's rule but the process cannot be stopped. Thank you.
 A: \begin{align}
l&=\lim_{w\rightarrow \Psi^{-1}(a)} \frac{1}{(\Psi(w)-a)(\Psi')^{1/2}(w)}+\frac{1}{(\Psi'(\Psi^{-1}(a)))^{3/2}(\Psi^{-1}(a)-w)}\\
&=\lim_{w\rightarrow \Psi^{-1}(a)} \frac{\frac1{w-\Psi^{-1}(a)}}{\frac{\Psi(w)-a}{w-\Psi^{-1}(a)}(\Psi')^{1/2}(w)}+\frac{1}{(\Psi'(\Psi^{-1}(a)))^{3/2}(\Psi^{-1}(a)-w)}\\
&=\lim_{w\rightarrow \Psi^{-1}(a)} \left(\frac{1}{\frac{\Psi(w)-a}{w-\Psi^{-1}(a)}(\Psi')^{1/2}(w)}\right)\frac1{w-\Psi^{-1}(a)}\left(1-\frac{\frac{\Psi(w)-a}{w-\Psi^{-1}(a)}(\Psi')^{1/2}(w)}{(\Psi'(\Psi^{-1}(a)))^{3/2}}\right)\\
&=\frac1{(\Psi'(\Psi^{-1}(a)))^{3/2}}\lim_{w\rightarrow \Psi^{-1}(a)} \left(\frac{1}{\frac{\Psi(w)-a}{w-\Psi^{-1}(a)}(\Psi')^{1/2}(w)}\right)\left(\frac{(\Psi'(\Psi^{-1}(a)))^{3/2}-\frac{\Psi(w)-a}{w-\Psi^{-1}(a)}(\Psi')^{1/2}(w)}{w-\Psi^{-1}(a)}\right)\\
\end{align}
Now define \begin{align}f(w)=\frac{\Psi(w)-a}{w-\Psi^{-1}(a)}(\Psi')^{1/2}(w)\end{align}
which helps to see that the second bit is (minus) the derivative of $f(w)$ at $w=\Psi^{-1}(a)$. The first bit is clear. To calculate the second bit use a Taylor's expansion of $\Psi(w)$ at $\Psi^{-1}(a)$ and simplify to obtain: $$l=\frac1{(\Psi'(\Psi^{-1}(a)))^{3/2}}\times \frac1{(\Psi'(\Psi^{-1}(a)))^{3/2}} \times \frac{-1}2 2\Psi''\left(\Psi^{-1}(a)\right)(\Psi'(\Psi^{-1}(a)))^{1/2}=\color{red}{-\frac{\Psi''\left(\Psi^{-1}(a)\right)}{(\Psi'(\Psi^{-1}(a)))^{5/2}}}.$$
A: We are given a holomorphic function $f$ defined in a neighborhood $U$ of $c\in{\mathbb C}$, and it is assumed that
$$f(c)=a,\qquad f'(w)=g^2(w)\quad(w\in U),\quad g(c)\ne0\ .$$
Put
$$\Phi(w):={1\over\bigl(f(w)-f(c)\bigr)g(w)}-{1\over g^3(c)(w-c)}={g^3(c)(w-c)-\bigl(f(w)-f(c)\bigr)g(w)\over g^3(c)g(w)\bigl(f(w)-f(c)\bigr)(w-c)}\ .\tag{1}$$ 
Now the numerator $N$ on the RHS can be developped as follows:
$$\eqalign{N&=g^3(c)(w-c)-\cr&\qquad\left(f'(c)(w-c)+{f''(c)\over2}(w-c)^2+?(w-c)^3\right)\bigr(g(c)+g'(c)(w-c)+?(w-c)^2\bigr)\cr
&=-\left({f''(c)g(c)\over2}+f'(c)g'(c)\right)(w-c)^2+?(w-c)^3\ .\cr}$$
By definition of $g$ we have $f''=2gg'$, so that $f'g'=f'{f''\over 2g}={f''g\over2}$. This allows to write $N$ as
$$N=-f''(c)g(c)(w-c)^2+?(w-c)^3\ .\tag{2}$$
The denominator $D$ on the RHS of $(1)$ is simpler:
$$D=g^3(c)g(w)\bigl(f'(c)(w-c)^2+?(w-c)^3)\ .\tag{3}$$
Plugging $(2)$ and $(3)$ into $(1)$ we obtain
$$\Phi(w)=-{f''(c)g(c)+?(w-c)\over g^3(c)g(w)\bigl(f'(c)+?(w-c)\bigr)}\ ,$$
so that
$$\lim_{w\to c}\Phi(w)=-{f''(c)g(c)\over g^4(c)f'(c)}=-{f''(c)\over\bigl(f'(c)\bigr)^{5/2}}\ .$$
(In the above the ? stands in for various functions that are analytic in $U$.)
