Proving a function is continuous for a fixed variable This is from Conway's book:
Suppse $f:G \rightarrow \mathbb{C}$ is analytic and define $\phi: G \times G \rightarrow \mathbb{C}$ by $\phi(z,w)=\frac{f(z)-f(w)}{(z-w)}$ if $z \neq w$ and $\phi(z,z)=f'(z)$.  Prove that $\phi$ is continuous...
Because $f$ is analytic, $f'$ exists for all points $z$ in the region $G \times G$.
I think I get for free that $f'$ is continuous.  But I'm not sure where to go from here.
 A: Let me show you why there is a subtlety here. Let's switch to the world of real functions. Let
$$f(x) = \begin{cases} x^2 \sin(x^{-1}) & x \neq 0 \\ 0 & x=0 \end{cases}$$
Then $f$ is differentiable. (Obvious at $x \neq 0$, and $\lim_{h \to 0} (f(h)-f(0))/h = \lim_{h \to 0} h \sin h^{-1}=0$.)
Now, let $x = ( (2N+1/2)\pi)^{-1}$ and $y = ((2N-1/2)\pi)^{-1}$ for $N$ an integer. So 
$$\frac{f(x) - f(y)}{x-y} = \frac{( (2N+1/2)\pi)^{-2} + ((2N-1/2)\pi)^{-2}}{( (2N+1/2)\pi)^{-1} - ((2N-1/2)\pi)^{-1}}$$
$$\approx \frac{1}{\pi} \frac{2 (2N)^{-2}}{- (2N)^{-2}} = - \frac{2}{\pi}.$$
So, as $(x,y) \to (0,0)$ along the sequence $(( (2N+1/2)\pi)^{-1}, ( (2N-1/2)\pi)^{-1})$, the ratio $(f(x)-f(y))/(x-y)$ does not go to $0$.
Your job is to show that this doesn't happen in the complex world.
A: The statement follows immediately from the equality
$$
f(z)-f(w)=(z-w)\ \int_0^1\ f'(w+tz-tw)\ dt.\tag1
$$
EDIT 1.  More precisely, the formula makes sense for $z$ and $w$ belonging to any given convex subset of $G$. But this is sufficient, because the continuity is obvious off the diagonal. 
EDIT 2. 


*

*The above argument shows that $$\frac{f(z)-f(w)}{z-w}$$ is holomorophic. 

*To prove $(1)$, set $g(t):=f(w+tz-tw)$ and observe $g(1)-g(0)=\int_0^1\,g'(t)\,dt$.
