Limit of $\frac{\exp(z)-1}{z}$ for $z \rightarrow 0$

Question

I would like to show that the limit of $\frac{\exp(z)-1}{z}$ for $z \rightarrow 0$ is $1$. But I don't know how to use the definition of $\exp(z)=\sum\limits_{n=0}^{\infty}\frac{z^n}{n!}$.

Answer 1

HINT

$$\exp(z)=1+\sum_{n=1}^{\infty}\frac{z^n}{n!}$$

Answer 2

$ \left | \frac{\exp(z)-1}{z}-1 \right | =\left| \frac{1}{z} \sum_{n=1}^{\infty}\frac{z^n}{n!}-1 \right |= \left | \sum_{n=0}^{\infty} \frac{z^n}{(n+1)!}-1\right|= \left|\sum_{n=1}^{\infty} \frac{z^n}{(n+1)!} \right|= \vert z \vert \left| \sum_{n=0}^{\infty} \frac{z^n}{(n+2)!}\right| ≤\vert z \vert \cdot e $

therefore you get then $\lim_{z \rightarrow0} \left|\frac{\exp(z)-1}{z}-1 \right|=0 \Longrightarrow \lim_{z \rightarrow0} \left|\frac{\exp(z)-1 }{z} \right|=1$