$\lim_{h\to 0} \frac{E(h)-1}{h}=1$ Let $E(z) = \sum_\limits{n=0}^{\infty} \dfrac{z^n}{n!}$.  In proving that $E'(z) = E(z),$ the author writes $$\lim_{h\to 0} \frac{E(h)-1}{h}=1,$$ claiming that this fact follows from the definition of $E(z)$.
How exactly does it follow from the definition of $E(z)$?
 A: Note that, from the series definition of $E(z)$, $E(z)=1+\color{red}z+\color{blue}{O(z^2)},$ so $\dfrac {E(z)-1}z=\color{red}1+\color{blue}{O(z)}.$
A: Basically one has that
$$\lim_{h\to 0} \frac{E(h)-1}{h}=\lim_{h\to 0} \frac{\sum_{n=0}\frac{h^n}
{n!}-1}{h}=\lim_{h\to 0} \frac{{1}+\sum_{n=1}\frac{h^n}
{n!}-1}{h}=\lim_{h\to 0}\sum_{n=1}\frac{h^{n-1}}
{n!}=1+\underbrace{\lim_{h\to 0}\sum_{n=2}\frac{\overbrace{h^{n-1}}^{\to 0}}
{n!}}_{=0}=1$$
where I've first used the definition, then taken out the first term, canceled, divided and lastly again taken out the first term again where the series tends to zero as its convergence radius is infinite and thus you can bring in the limit.
A: Without using the Taylor series, as this may lead to a closed loop in the reasoning (depending on how you define function $E$):
Using Bernoulli inequality you have $$ (1+\frac{x}{n})^n \ge 1+ x \qquad \text{for }x\ge -n $$
Therefore:
$$ E(x) = \lim_{n\rightarrow\infty} (1+\frac{x}{n})^n \ge 1+ x \qquad \text{for }x \in \mathbb{R} $$
By taking the inverse for $x>-1$:
$$ E(-x) \le \frac{1}{1+x} \qquad \text{for }x> -1 $$
$$ E(x) \le \frac{1}{1-x} \qquad \text{for }x< 1 $$
We have then
$$ 1+x \le E(x) \le \frac{1}{1-x} \qquad \text{for }x< 1 $$
$$ x \le E(x) - 1\le \frac{x}{1-x} \qquad \text{for }x< 1 $$
$$ 1 \le \frac{E(x) - 1}{x} \le \frac{1}{1-x} \qquad \text{for }x\in(0,1) $$
$$ \frac{1}{1-x} \le \frac{E(x) - 1}{x} \le 1 \qquad \text{for }x<0 $$
From the comparison of the limits we get $$\lim_{x\rightarrow 0} \frac{E(x) - 1}{x} = 1$$
