Proving $\lim_{h\to 0} \dfrac{e^h - 1}{h} = 1$ when differentiating $e^x$ by first principles

Question

The definition of $e$ as continual compounding is found computationally as: $\lim_{n\to \infty} \Big(1+\dfrac{1}{n}\Big)^n = e$

If we let $h=\dfrac{1}{n}$, then $e = \lim_{h\to 0} (1+h)^{\frac{1}{h}}$

Differentiating $e^x$ by first principles we get:

$f(x)=e^x$

$f'(x)=\lim_{h\to 0} \dfrac{e^{x+h}-e^x}{h}$

$=\lim_{h\to 0} \dfrac{e^x\times e^h-e^x}{h}$

$=\lim_{h\to 0} \dfrac{e^x(e^h - 1)}{h}$

$=e^x \times \lim_{h\to 0} \dfrac{e^h - 1}{h}$

Now, I know the result of the limit is going to be $1$. So my intuition is that I should be able to substitute in the compounding expression for $e$ and then manipulate the limit in the derivative to yield $1$. So by reverse engineering the expected solution I have the following:

$\lim_{h\to 0} \dfrac{e^h - 1}{h} = \lim_{h\to 0} \dfrac{[\lim_{h\to 0} (1+h)^{\frac{1}{h}}]^h - 1}{h}$

$= \lim_{h\to 0} \dfrac{[(1+h)^\frac{1}{h}]^h -1}{h}$

$= \lim_{h\to 0} \dfrac{(1+h)-1}{h}$

$= \lim_{h\to 0} \dfrac{h}{h}$

$= \lim_{h\to 0} 1$

$= 1$

My question is regarding this step:

$\lim_{h\to 0} \dfrac{[\lim_{h\to 0} (1+h)^{\frac{1}{h}}]^h - 1}{h} = \lim_{h\to 0} \dfrac{[(1+h)^\frac{1}{h}]^h -1}{h}$

I believe it must be true since it produces the expected result, but I'm unsure how to justify removing the nested limit.

Answer 1

If you know $(e^x)'=e^x$, then by the definition of the derivative, $$1=e^0=(e^x)'(0)=\lim_{h\to0}\dfrac {e^h-e^0}h=\lim_{h\to0}\dfrac {e^h-1}h$$.

But I guess that begs the question.

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There's L'hopital. But it also requires knowing that derivative:

$$\lim_{h\to0}\dfrac {e^h-1}h=\lim_{h\to0}\dfrac {e^h}1=e^0=1$$

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If you know the power series $$e^x=\sum_n\dfrac {x^n}{n!}$$. Then it's pretty easy to see that $(e^x)'=e^x$.

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Or, you could use the fact that $e^x$ and $\ln x$ are inverses. If you know $(\ln x)'=1/x$, then since $e^{\ln x}=x$, you get by the chain rule that $(e^x)'(\ln x)\cdot 1/x=1$. So $(e^x)'(\ln x)=x$. And $(e^x)'$ is $\ln^{-1} x=e^x$.

Answer 2

You may also be interested in the following (equivalent) definition of the exponential function: \begin{align*} \exp(x) = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \ldots = \sum_{n=0}^{\infty}\frac{x^{n}}{n!} \end{align*}

Such power series converges for every possible value of $x\in\mathbb{R}$. Hence we conclude that $e^{h} - 1 = h + o(h)$.

Having said that, the proposed limit can be computed in the following manner: \begin{align*} \lim_{h\to 0}\frac{e^{h} - 1}{h} = \lim_{h\to 0}\frac{h + o(h)}{h} = \lim_{h\to 0}\left(1 + \frac{o(h)}{h}\right) = 1 \end{align*}

EDIT

This is particularly useful to obtain the derivative of $e^{x}$ by the limit definition: \begin{align*} (e^{x})' = \lim_{h\to 0}\frac{e^{x + h} - e^{x}}{h} = \lim_{h\to 0}\frac{e^{x}(e^{h} - 1)}{h} = e^{x} \end{align*}

and we are done.

Hopefully this helps!

Answer 3

How is $b^x; b > 0; x\in \mathbb R$ defined?

If we define $b^x = \lim_{q_i\to x;q_i\in \mathbb Q} b^q$ where $\{q_i\}$ is any sequence of rationals so that $q_i \to x$, then we have the rather ... surreal:

$\lim_{h\to 0}\frac {e^h -1}{h} = \lim_{h\to 0}\lim_{q_i\to h}\frac {\lim_{n\to\infty} (1+\frac 1n)^n)^h-1}{q_{h,i}}$

where for each $h$ we have $q_{h,i}$ be a sequence of rational that converge to that $h$.

Now we can, as we are taking limits, take any sub limits we like if that will make things simpler. So.... bear with me.... As you know, we can always replace a limit of $\lim_{h\to 0}f(h)$ with $\lim_{n\to \infty} f(\frac 1n)$. And if we restrict $n\in \mathbb Z$ we have $\frac 1n \in \mathbb Q$ so we can let the nescessary $q_{h,i}\to h$ all be equal to $\frac 1n$ as the constant sequence $\frac 1n,\frac 1n, \frac 1n....$ certainly does converge to $\frac 1n$.

So more surreal but ultimately... hey, that's legit! we have

$\lim_{h\to 0}\frac {e^h - 1}h =$
$\lim_{h=\frac 1n; q_i=h=\frac 1n;i\to \infty; n\to\infty}\frac {((1+\frac 1n)^n)^{q_i}-1}h=$
$\lim_{n\to \infty}\frac {((1+\frac 1n)^n)^{\frac 1n} - 1}{\frac 1n}=$
$\lim_{n\to \infty}n\cdot ((1+\frac 1n)-1)=\lim_{n\to \infty}n\cdot \frac 1n = \lim_{n\to\infty}1=1$

Now... that assumes we have defined $b^x;b> 0; x\in \mathbb R$ as $\lim_{q_i\to x}b^{q_i}$

Most calculus books don't.

More standard is to define $\ln x = \int_0^x \frac 1t dt$ and to define $exp(x)$ to be the inverse of this. That is $exp(x)=y \iff \int_1^y\frac 1t dt = x$. And define $b^x = exp(x\cdot \ln b)$ and $e=exp(1)$ or in other words $e$ is the positive real value where $\int_1^e\frac 1t dt = 1$.

In that case $\lim_{h\to 0}\frac {b^h -1}h=\ln b$ sort of falls into place by definition.

As does proving that $\lim_{n\to \infty}(1+\frac 1n)^n=e$.

(See https://www2.clarku.edu/faculty/djoyce/ma121/elimit.pdf)