I'm looking for a straight forward proof using the definition of a derivative applied to the exponential function and substitution of one of the limit definitions of $e$, starting with
$e = \lim_{h\to \infty}\left({1+\dfrac{1}{h}}\right)^h$ or $e=\sum_{h=0}^{\infty}{\dfrac{1}{h!}}$
and
$\dfrac{d}{dx}\left( e^x \right) = \lim_{h\to 0}\left({\dfrac{e^{x+h}-e^{x}}{h}}\right)$
I found a proof I sort of liked here (which is sort of along the lines of a proof I'd like to use):
My only problem is that he combines the dummy variable, $h$, for the limit definition of $e$ and the dummy variable, $h$, used for the derivative. To me, it seems like it's not quite valid to do such a thing because it assumes both values are equal. Can anyone provide a better proof or justification for why the dummy variables can be combined?
EDIT:
I guess I'd also like to have a proof of why:
$\lim_{h\to 0}\left({\dfrac{e^{h}-1}{h}}\right) = 1$
using one of the limit definitions of $e$ shown above.
$\endgroup$ 84 Answers
$\begingroup$As I can see from your edit, you noticed that the only difficult part is to show that $\exp'(0) = 1$, that is, $\lim_{h\to 0}\left({\dfrac{\exp(h)-1}{h}}\right) = 1$.
So here's my proof, using only the definition of the exponential function and elementary properties of limits.
We use the following definition of the exponential function:
\begin{align*} &\exp : \mathbb{R} \to \mathbb{R}\\ &\exp(x) = \lim_{k \to +\infty} \left(1 + \frac{x}{k}\right)^k \end{align*}
Let's define \begin{align*} &A : \mathbb{R}^* \to \mathbb{R}\\ &A(h) = \frac{\exp(h) - 1}{h} - 1 \end{align*}
We're going to show that $\lim_{h \to 0} A(h) = 0$. This will imply that $\lim_{h \to 0} \frac{\exp(h) - 1}{h} = 1$ and consequently, that $\exp'(0) = 1$.
Let's show that for all $h \in [-1,1]\setminus\{0\}$, $|A(h)| \leq |h|$
Let $h \in [-1,1]\setminus\{0\}$. We define the sequence $(u_k)_{k \in \mathbb{N}^*}$ by \begin{align*} u_k = \frac{\left(1+\frac{h}{k}\right)^k - 1}{h} - 1 \end{align*}
From the definition of the exponential function and from the rules of addition and multiplication of limits, we get:
\begin{align*} \lim_{k \to + \infty} u_k = A(h) \end{align*}
The continuity of the absolute value function then brings:
\begin{align*} \lim_{k \to + \infty} |u_k| = |\lim_{k \to + \infty} u_k| = |A(h)| \end{align*}
If we manage to show that after a certain rank, $|u_k| \leq |h|$, we'll be able to conclude that $|A(h)| = \lim_{k \to +\infty}|u_k| \leq |h|$.
For $k \in \mathbb{N}^*$, we have
$$ u_k = \frac{\left(\sum\limits_{i=0}^{k} \binom{k}{i} \left(\frac{h}{k}\right)^i \right) - 1 - h}{h} = \frac{1}{h}\sum_{i=2}^k \frac{\binom{k}{i}}{k^i} h^i = h \sum_{i=2}^k \frac{\binom{k}{i}}{k^i} h^{i-2} $$
The triangle inequality brings: $$ |u_k| \leq |h| \sum_{i=2}^k \frac{\binom{k}{i}}{k^i} |h|^{i-2} $$
We have $h \in [-1,1]$. So $|h|^{i-2} \leq 1$ for every $i \in \mathbb{N}$ such as $i \geq 2$. Moreover, for $k,i \in \mathbb{N}\setminus\{0,1\}$: $$ \frac{\binom{k}{i}}{k^i} = \frac{\prod\limits_{j=0}^{i-1}(k-j)}{i!\prod\limits_{j=0}^{i-1}k} \leq \frac{1}{i!} \leq \frac{1}{2^{i-1}} $$ Therefore, as soon as $k \geq 2$: $$ |u_k| \leq |h| \sum_{i=2}^k \frac{1}{2^{i-1}} = |h| \sum_{i=1}^{k-1} \frac{1}{2^i} = |h| \left(\frac{1-\frac{1}{2^k}}{1-\frac{1}{2}} - 1\right) = |h| \left(1-\frac{1}{2^{k-1}}\right) \leq |h| $$
Hence: $$ |A(h)| = \lim_{k \to \infty} |u_k| \leq |h| $$ This is true for all $h \in [-1,1]\setminus\{0\}$. Therefore: $$ \lim_{h \to 0} A(h) = 0 $$ which shows that $\exp'(0) = 1$.
$\endgroup$ $\begingroup$$$\frac{e^h-1}h=\frac1h\left(h+\frac{h^2}{2!}+\frac{h^3}{3!}+\ldots\right)=1+\frac h{2!}+\frac{h^2}{3!}+\ldots\xrightarrow[h\to 0]{}1$$
Of course, some power series theory must be known to fully justify the above. And now all it's easy:
$$\lim_{h\to 0}\frac{e^{x+h}-e^x}h=\lim_{h\to 0}\,e^x\frac{e^h-1}h=e^x\cdot 1=e^x$$
$\endgroup$ 2 $\begingroup$In regards to why $$\lim_{h\to\infty}\left(\frac{e^h-1}{h}\right)=1,$$ this really can be taken as a definition of $e$ to see why consider the derivative of a generic exponential function $a^x$: $$\frac{\mathrm{d}}{\mathrm{d}x}a^x=\lim_{h\to0}\frac{a^{x+h}-a^x}{h}=\lim_{h\to0}a^x\left(\frac{a^h-1}{h}\right)=a^x\lim_{h\to0}\left(\frac{a^h-1}{h}\right),$$ so $\frac{\mathrm{d}}{\mathrm{d}x}a^x\propto a^x$ and if we want this to be an equality we must require that the function $$f(x)=\lim_{h\to0}\left(\frac{x^h-1}{h}\right),$$ is $1$ for some $x$ $\big($incidentally $f(x)=\ln(x)\big)$.
If we then make the substitution $h\mapsto 1/k$ then $h\to0⁺$ as $k\to\infty$ so if we just forget about the other limit as $h\to0⁻$ for now, we write $$f_k(x)=k\left(\sqrt[k]{x}-1\right).$$ So we want to find an $x$, such that $f(x)=\lim_{k\to\infty}f_k(x)=1$ and from the expression for $f_k(x)$ it should be pretty obvious that for each $k$ there is a number $x_k$ such that $f_k(x_k)=1$, namely take $$x_k=\left(1+\frac{1}{k}\right)^k$$ then $$f_k(x_k)=k\left(\Bigg(\left(1+\frac{1}{k}\right)^k\Bigg)^{1/k}-1\right)=k\left(1+\frac{1}{k}-1\right)=1$$ so now $f_k(x_k)=1,\;\forall k$ in other words $\{f_k(x_k)\}$ is a constant sequence with all entries equal to $1$ so we have $$\lim_{k\to\infty}f_k(x_k)=\lim_{k\to\infty}f(x_k)=f\left(\lim_{k\to\infty}x_k\right)=1,$$ since firstly $f(x)=\lim_{k\to\infty}f_k(x)$ and the last step is justified since it turns out that $f(x)$ is continuous.
So by defining $$e=\lim_{k\to\infty}x_k=\lim_{k\to\infty}\left(1+\frac{1}{k}\right)^k$$ we have that $$\lim_{h\to\infty}\left(\frac{e^h-1}{h}\right)=1,$$ which in turn means that $$\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x.$$
$\endgroup$ $\begingroup$Proof for $\lim_{h\to 0}\left({\dfrac{e^{h}-1}{h}}\right) = 1$
Proof:
$e = \lim_{n\to \infty}\left({1+\frac{1}{n}}\right)^n$
let $h=\frac{1}{n}$
$e = \lim_{h\to 0}\left({1+h}\right)^\frac{1}{h}$
$\lim_{h\to 0}e^h=\lim_{h\to 0}(1+h)$
$\lim_{h\to 0}\left({\frac{1+h-1}{h}}\right) = 1$
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