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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):

http://www.math.brown.edu/UTRA/explog.html

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.

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  • $\begingroup$ For the last limit, you may use the definition of $e$ you wrote first, i.e. $e^h=\sum_{k\geq 0} \frac{h^k}{k!}$. $\endgroup$
    – Martingalo
    Commented Feb 10, 2014 at 19:36
  • $\begingroup$ @Martingalo I'd prefer not to go that route. $\endgroup$ Commented Feb 10, 2014 at 19:38
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    $\begingroup$ The first definition of $\;e\;$ requires $\;h\to \infty\;$ , not to zero. $\endgroup$
    – DonAntonio
    Commented Feb 10, 2014 at 19:38
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    $\begingroup$ What is your definition of the function $x \mapsto e^x$? $\endgroup$
    – Ulrik
    Commented Feb 10, 2014 at 19:39
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    $\begingroup$ The comment by @Svinepels is to the point: Defining $e^x$ is the hard part. The common approaches to doing so have the derivative formula as an easy corollary. $\endgroup$ Commented Feb 10, 2014 at 19:41

4 Answers 4

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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 $\exp'(0) = 1$.

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

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

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

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    $\begingroup$ I thinks it is kind of problematically, because to write the power series of $e^x$ you need to know how to derive it, and exactly this you want to find out. $\endgroup$
    – Victor
    Commented Aug 8, 2015 at 8:59
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    $\begingroup$ Can you explain please why $1+\frac{h}{2!}+\frac{h^2}{3!}+\cdots \to 1$ as $h\to 0$? $\endgroup$
    – RFZ
    Commented Dec 15, 2015 at 19:51
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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 $$\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x.$$

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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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  • $\begingroup$ well this shows that $\lim_{n\to \infty}\big({\dfrac{e^{1/n}-1}{1/n}}\big) = 1$ and not $\lim_{h\to 0}\big({\dfrac{e^{h}-1}{h}}\big) = 1$. $\endgroup$
    – Surb
    Commented Jun 6, 2017 at 7:10
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    $\begingroup$ Correcting the previous comment, this shows that $\lim_{h\rightarrow 0} \text{e}^h=1$ and not $\lim_{h\rightarrow 0} \frac{\text{e}^h-1}{h}=1$. $\endgroup$
    – Marra
    Commented Apr 4, 2019 at 19:30

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