Derivative of exponential function proof

Solution 1:

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 $\exp'(0) = 1$.

Solution 2:

$$\frac{e^h-1}h=\frac1h\left(h+\frac{h^2}{2!}+\frac{h^3}{3!}+\ldots\right)=1+\frac h{2!}+\frac{h^2}{3!}+\ldots\xrightarrow[h\to 0]{}1$$

Of course, some power series theory must be known to fully justify the above. And now all it's easy:

$$\lim_{h\to 0}\frac{e^{x+h}-e^x}h=\lim_{h\to 0}\,e^x\frac{e^h-1}h=e^x\cdot 1=e^x$$