question about derivative of exponential function When I proved derivation the exponential function expose with problem that have to use derivative of $e^x$
$$\frac{de^x}{dx} = \lim_{h\to 0}\frac{e^{x+h} -e^x}h=\lim_{h\to 0} e^x  \frac{e^h-1}h =e^x \cdot \lim_{h\to 0}  \frac{e^h-1}h$$
Calculate  $\displaystyle\lim_{h\to 0}  \frac{e^h-1}h$ but can’t use l’hopital theorem and Taylors theorem because use derivative of $e^x$ .
Please help me to solve it.
 A: I assume, you can use that $\ e^h=\displaystyle\lim_{n\to\infty}\left(1+\frac hn \right)^n$.
Hint: Use the Bernoulli inequation: $(1+x)^{\alpha} \ge 1+\alpha x$ if $x > -1$ and $\alpha>0$, so it yields $e^h\ge 1+h\ $ if $\ h> -1$, and take its reciprocal for the converse to prove that the limit you look for is $1$.
A: Sometimes the fact that that limit is $1$ is taken to be the definition of $e$.  (That's what's done in Stewart's textbook.)
Supposing you had $\dfrac{d}{dx} 4^x$ instead of $\dfrac{d}{dx} e^x$.  You'd get $4^x\cdot\lim\limits_{h\to0}\dfrac{4^h - 1}{h}$.
But the function $y=4^x$ gets steeper as you go from left to right, and the slope of its secant line through the points where $x=-1/2$ and $x=0$ is $1$; therefore the slope of the curve at $x=0$ is more than $1$.  If you had $y=2^x$, you'd consider the secant line at $x=0$ and $x=1$ and conclude that the slope at $0$ is less than $1$.
So $4$ is too big, and $2$ is too small, to be the base of the natural exponential function.  Somewhere between $2$ and $4$ is the right number.  If $e$ is that right number, then of course the slope at $0$ is $1$.
(What this omits is how we know just where that number is, i.e. that it's $2.71828\ldots$, beyond the fact that it's between $2$ and $4$.)
A: Use the definition of limit: $$\lim _{ h\rightarrow 0 }{ \frac { { e }^{ x+h }-{ e }^{ x } }{ h }  } ={ e }^{ x }\lim _{ h\rightarrow 0 }{ \frac { { e }^{ h }-1 }{ h }  } ={ e }^{ x }\lim _{ h\rightarrow 0 }{ \frac { { e }^{ h }-{ e }^{ 0 } }{ h }  } .$$
$\lim _{ h\rightarrow 0 }{ \frac { { e }^{ h }-{ e }^{ 0 } }{ h }  } $ is the derivative of $e^x$ at $x=0$. Therefore we can write $\frac { d }{ dx } { e }^{ x }$ as $$\frac { d }{ dx } { e }^{ x }={ e }^{ x }{ \left[ \frac { d }{ dx } { e }^{ x } \right]  }_{ x=0 }.$$ Open ${ \left[ \frac { d }{ dx } { e }^{ x } \right]  }_{ x=0 }$ in a similar way, you get $${ \left[ \frac { d }{ dx } { e }^{ x } \right]  }_{ x=0 }={ e }^{ 0 }{ \left[ \frac { d }{ dx } { e }^{ x } \right]  }_{ x=0 }={ e }^{ 0 }{ e }^{ 0 }{ \left[ \frac { d }{ dx } { e }^{ x } \right]  }_{ x=0 }=1\cdot 1\cdot 1\cdot 1\cdot 1\cdot 1\cdots .$$ As a result $$\frac { d }{ dx } { e }^{ x }={ e }^{ x }.$$
