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).
 A: 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 $0$.
A: Let us define the exponential using integer powers by
$$e^x=\lim_{n\to\infty}(1+\frac xn)^n.$$
This definition is consistent with that of $e$, 
$$e=\lim_{n\to\infty}(1+\frac1n)^n=\lim_{h\to0}(1+h)^{1/h}.$$
Then, using the ordinary binomial theorem
$$(1+\frac xn)^n
=\sum_{k=0}^n\binom nk\left(\frac xn\right)^k
=1+\frac nnx+\frac{n(n-1)}{2n^2}x^2+\frac{n(n-1)(n-2)}{2.3.n^3}x^3+\cdots+\frac{x^n}{n^n}.$$
All coefficients are positive and bounded by $1$, hence
$$1+x\le\left(1+\frac xn\right)^n\le\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}\le\frac1{1-x},$$
for $0\le x<1$.
In the limit $n\to\infty$,
$$1+x\le e^x\le\frac1{1-x}.$$
Then
$$1\le\frac{e^x-1}x\le\frac1{1-x},$$
and the limit for $x\to0$ is clearly $1$. Similar bracketing holds for $x<0$.
A: Note you can view $\displaystyle \lim_{h\to 0} \dfrac{e^h-1}{h} = (e^h)'(0)=1$
