$\lim_{h\rightarrow 0} \dfrac {e^{f(z+h)}-e^{f(z)}}{f(z+h)- f(z)}$ given that $V$ is an open subset of $\mathbb{C}$ and $z \in V$, calculate $\lim_{h\rightarrow 0} \dfrac  {e^{f(z+h)}-e^{f(z)}}{f(z+h)- f(z)}$, if $f$ is known to be a continuous complex function in $V$. I know that the result is supposed to be  $e^{f(z)}$, but I can't figure out the details in a rigorous way. Any help appreciated and many thanks in advance.
 A: Hint: Write $w_0 = f(z)$ and $w = f(z+h)$. Note that the continuity of $f$ implies
$$(h\to 0) \Rightarrow (w \to w_0).$$
The continuity of $f$ in $z$ means that for any arbitrarily prescribed $\delta > 0$, we can find an $\eta > 0$ such that we have $\lvert f(z+h) - f(z)\rvert < \delta$ for all $h$ with $\lvert h\rvert < \eta$.
We have - if $f(z+h) \neq f(z)$ -
$$
\left\lvert \frac{e^{f(z+h)}- e^{f(z)}}{f(z+h)-f(z)} - e^{f(z)}\right\rvert
= \left\lvert e^{f(z)}\right\rvert\cdot\left\lvert \frac{e^{f(z+h)-f(z)}-1}{f(z+h)-f(z)}-1\right\rvert.
$$
Let us for brevity write $y = f(z+h)-f(z)$. 
For $\lvert y\rvert < \frac12$, we have
$$\lvert e^y - 1-y\rvert = \left\lvert \sum_{n=2}^\infty \frac{y^n}{n!}\right\rvert \leqslant \sum_{n=2}^\infty \frac{\lvert y\rvert^n}{n!} \leqslant \frac{\lvert y\rvert^2}{2}\sum_{k=0}^\infty \lvert y\rvert^k = \frac{\lvert y\rvert^2}{2(1-\lvert y\rvert)} < \lvert y\rvert^2,$$
and therefore
$$
\left\lvert \frac{e^y-1}{y} - 1\right\rvert = \left\lvert \frac{e^y-1-y}{y}\right\rvert \leqslant \lvert y\rvert.
$$
Now, given an arbitrary $\varepsilon > 0$, we set $\delta = \min \left\lbrace \frac12, \frac{\varepsilon}{\lvert e^{f(z)}\rvert}, \varepsilon\right\rbrace$, and for that $\delta$ choose $\eta > 0$ such that $\lvert f(z+h)-f(z)\rvert < \delta$ for $\lvert h\rvert < \eta$. Then we have
$$\left\lvert \frac{e^{f(z+h)}- e^{f(z)}}{f(z+h)-f(z)} - e^{f(z)}\right\rvert \leqslant \lvert e^{f(z)}\rvert \cdot \lvert f(z+h) - f(z)\rvert < \lvert e^{f(z)}\rvert\cdot\delta < \varepsilon$$
for $\lvert h\rvert < \eta$, so
$$\lim_{h\to 0} \frac{e^{f(z+h)}-e^{f(z)}}{f(z+h)-f(z)} = e^{f(z)},$$
subject to the restriction that we must have $f(z+h) \neq f(z)$.
A: HINT. Expand both numerator and denominator as Taylor series around h=0. Limit the expansion to first order, it will be sufficient. Can you continue from here ? All of this supposes that f(z) is differentiable.
A: This is much easier than it seems. Don't be thrown off by the usage of $f(z + h) - f(z)$, which hints at dividing by $h$. The only thing that is important is that $f(z + h) \to f(z)$ when $h \to 0$ and the definition of the derivative of the exponential function.
Concretely, use the following general fact, which follows straightforwardly from the $\epsilon$-$\delta$-definition of limits.
Theorem. If $\lim_{x \to x_0} a(x) = l$ and $\lim_{h \to h_0} b(h) = x_0$, then $\lim_{h \to h_0} a(b(h)) = l$.
Now apply this to $a(x) := \frac{\exp(x) - \exp(f(z))}{x - f(z)}$ with $x \to f(z)$ and $b(h) := f(z + h)$ with $h \to 0$. Note that $\lim_{h \to 0} b(h) = f(z)$ because $f$ is continuous and that $\lim_{x \to f(z)} a(x) = \exp'(f(z))$ by definition of $\exp'$, so $\lim_{x \to f(z)} a(x) = \exp(f(z))$. Consequently, using the theorem above, $\lim_{h \to 0} \frac{\exp(f(z + h)) - \exp(f(z))}{f(z+h) - f(z)} = \lim_{h \to 0} a(b(h)) = \exp(f(z))$.
