Using the definition to show that $D_{x}(e^x)=e^x$. Our applied calculus text defines $e$ by $e=\displaystyle\lim_{h\to0}(1+h)^{1/h}$,  
and then gives the following argument to show that $D_{x}(e^x)=e^x$:
If $f(x)=e^x$, then 
$\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{e^{x+h}-e^x}{h}=\lim_{h\to0}\frac{e^x(e^h)-e^x}{h}=\lim_{h\to0}\frac{e^x(1+h)-e^x}{h}=\lim_{h\to0}\frac{e^x(h)}{h}=e^x.$

I have two questions about this argument:
1) Is there a simple way to justify replacing $e^h$ by $1+h$ in the numerator?*
2) If not, is there a way to prove that $\displaystyle \lim_{h\to0}\frac{e^h-1}{h}=0$ using the definition of $e$ given above?*
*(without using Taylor series or logarithms, neither of which have been covered at this point in the book).

We are going to show that 
$$\lim_{h \to 0} (1+h)^\frac{1}{h}=e$$
implies 
$$\lim_{t \to 0} \frac{e^{t}-1}{t}=1$$
Define $h_t=e^{t}-1$. When $t \to 0$ we have $h_t \to 0$. Therefore
$$\lim_{h_t \to 0} (1+h_t)^\frac{1}{h_t}=e$$
$$\lim_{t \to 0} (e^{t})^\frac{1}{e^{t}-1}=e$$
$$\lim_{t \to 0} e^\frac{t}{e^{t}-1}=e$$
Note that in general $\lim_{t \to a} f(g(t))=f(c)$ doesn't imply that $\lim_{t \to a} g(t)=c$. BUT, in this case the claim can be proven using the fact that the exponential is strictly increasing and continuous at $0$.
