TITLE: Using the Limit definition to find the derivative of $e^x$
QUESTION [35 upvotes]: I was wondering how we could use the limit definition 
$$ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$
to find the derivative of $e^x$, I get to a point where I do not know how to simplify the indeterminate $\frac{0}{0}$. Below is what I have already done
$$\begin{align}
&\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
&\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} \\
&\lim_{h \rightarrow 0} \frac{e^x (e^h-1)}{h} \\
&e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h}
\end{align}$$
Where can I go from here? Because, the $\lim$ portion reduces to indeterminate when $0$ is subbed into $h$.

REPLY [3 votes]: The discovery of the constant $e$ is credited to Jacob Bernoulli in 1683 who attempted to find the value of the following expression (which is equal to $e$):
$$\lim_{n\to \infty}{\left(1+\frac1n\right)}^n.$$
Alternatively, we can substitute $n=\frac1h$ to obtain:
$$e=\lim_{h\to0}(1+h)^{1/h}.$$
Substitute this limit into your expression to get:
$$e^x\lim_{h\to0}\frac{{\left((1+h)^{1/h}\right)}^h-1}h$$
$$=e^x\lim_{h\to0}\frac{{(1+h)}-1}h$$
$$=e^x.$$