29
$\begingroup$

I was wondering how we could use the limit definition

$$ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$

to find the derivative of $e^x$, I get to a point where I do not know how to simplify the indeterminate $\frac{0}{0}$. Below is what I have already done

$$\begin{align} &\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} \\ &\lim_{h \rightarrow 0} \frac{e^x (e^h-1)}{h} \\ &e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \end{align}$$

Where can I go from here? Because, the $\lim$ portion reduces to indeterminate when $0$ is subbed into $h$.

$\endgroup$
  • 1
    $\begingroup$ How do you define $e$? $\endgroup$ – Adam Saltz Apr 12 '13 at 2:22
  • $\begingroup$ @AdamSaltz $e$ as in euler's constant? $\endgroup$ – Jeel Shah Apr 12 '13 at 2:23
  • $\begingroup$ @julien Sorry, I don't understand what you mean by "the function of $e^x$". Can you please clarify? $\endgroup$ – Jeel Shah Apr 12 '13 at 2:24
  • $\begingroup$ @julien I don't think I understand what it means to define a function. Do you mean if it is an exponential function or a one to one function? $\endgroup$ – Jeel Shah Apr 12 '13 at 2:27
  • $\begingroup$ Some people define the exponential $e^x$ as the unique solution of the ode: $y'=y$ and $y(0)=1$. For these people, no needto prove that $\frac{d}{dx}e^x=e^x$. It holds by definition. Now given your title, there are I think two possibilities: $e^x=\lim_n \sum_{k=0}^n\frac{x^k}{k!}$, or $e^x=\lim_n\left(1+\frac{x}{n}\right)^n$. $\endgroup$ – Julien Apr 12 '13 at 2:36
37
$\begingroup$

Sometimes one defines $e$ as the (unique) number for which $$\tag 1 \lim_{h\to 0}\frac{e^h-1}{h}=1$$

In fact, there are two possible directions.

$(i)$ Start with the logarithm. You'll find out it is continuous monotone increasing on $\Bbb R_{>0}$, and it's range is $\Bbb R$. It follows $\log x=1$ for some $x$. We define this (unique) $x$ to be $e$. Some elementary properties will pop up, and one will be $$\tag 2 \lim\limits_{x\to 0}\frac{\log(1+x)}{x}=1$$

Upon defining $\exp x$ as the inverse of the logarithm, and after some rules, we will get to defining exponentiation of $a>0\in \Bbb R$ as $$a^x:=\exp(x\log a)$$

In said case, $e^x=\exp(x)$, as we expected. $(1)$ will then be an immediate consequence of $(2)$.

$(ii)$ We might define $$e=\sum_{k=0}^\infty \frac 1 {k!}$$ (or the equivalent Bernoulli limit). Then, we may define $$\exp x=\sum_{k=0}^\infty \frac{x^k}{k!}$$ Note $$\tag 3 \exp 1=e$$

We define the $\log$ as the inverse of the exponential function. We may derive certain properties of $\exp x$. The most important ones would be $$\exp(x+y)=\exp x\exp y$$ $$\exp'=\exp$$ $$\exp 0 =1$$

In particular, we have that $\log e=1$ by. We might then define general exponentiation yet again by $$a^x:=\exp(x\log a)$$

Note then that again $e^x=\exp x$. We can prove $(1)$ easily recurring to the series expansion we used.


ADD As for the definition of the logarithm, there are a few ones. One is $$\log x=\int_1^x \frac{dt}{t}$$

Having defined exponentiation of real numbers using rationals by $$a^x=\sup\{a^r:r\in\Bbb Q\wedge r<x\}$$

we might also define $$\log x=\lim_{k\to 0}\frac{x^k-1}{k}$$

In any case, you should be able to prove that

$$\tag 1 \log xy = \log x +\log y $$ $$\tag 2 \log x^a = a\log x $$ $$\tag 3 1-\dfrac 1 x\leq\log x \leq x-1 $$ $$\tag 4\lim\limits_{x\to 0}\dfrac{\log(1+x)}{x}=1 $$ $$\tag 5\dfrac{d}{dx}\log x = \dfrac 1 x$$

What you want is a direct consequence of either $(4)$ or $(5)$, or of the first sentence in my post.


ADD We can prove that for $x \geq 0$ $$\lim\left(1+\frac xn\right)^n=\exp x$$ from definition $(ii)$.

First, note that $${n\choose k}\frac 1{n^k}=\frac{1}{{k!}}\frac{{n\left( {n - 1} \right) \cdots \left( {n - k + 1} \right)}}{{{n^k}}} = \frac{1}{{k!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{k - 1}}{n}} \right)$$

Since all the factors to the rightmost are $\leq 1$, we can claim $${n\choose k}\frac{1}{{{n^k}}} \leqslant \frac{1}{{k!}}$$

It follows that $${\left( {1 + \frac{x}{n}} \right)^n}=\sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}} \leqslant \sum\limits_{k = 0}^n {\frac{{{x^k}}}{{k!}}} $$

It follows that if the limit on the left exists, $$\lim {\left( {1 + \frac{x}{n}} \right)^n} \leqslant \lim \sum\limits_{k = 0}^n {\frac{{{x^k}}}{{k!}}} = \exp x$$

Note that the sums in $$\sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}} $$

are always increasing, which means that for $m\leq n$

$$\sum\limits_{k = 0}^m {{n\choose k}\frac{{{x^k}}}{{{n^k}}}}\leq \sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}}$$

By letting $n\to\infty$, since $m$ is fixed on the left side, and $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{k!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{k - 1}}{n}} \right) = \frac{1}{{k!}}$$

we see that if the limit exists, then for each $m$, we have $$\sum\limits_{k = 0}^m {\frac{{{x^k}}}{{k!}}} \leqslant \lim {\left( {1 + \frac{x}{n}} \right)^n}$$

But then, taking $m\to\infty$ $$\exp x = \mathop {\lim }\limits_{m \to \infty } \sum\limits_{k = 0}^m {\frac{{{x^k}}}{{k!}}} \leqslant \lim {\left( {1 + \frac{x}{n}} \right)^n}$$

It follows that if the limit exists $$\eqalign{ & \exp x \leqslant \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n} \cr & \exp x \geqslant \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n} \cr}$$ which means $$\exp x = \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n}$$ Can you show the limit exists?

The case $x<0$ follows now from $$\displaylines{ {\left( {1 - \frac{x}{n}} \right)^{ - n}} = {\left( {\frac{n}{{n - x}}} \right)^n} \cr = {\left( {\frac{{n - x + x}}{{n - x}}} \right)^n} \cr = {\left( {1 + \frac{x}{{n - x}}} \right)^n} \cr} $$

using the squeeze theorem with $\lfloor n-x\rfloor$, $\lceil n-x\rceil$, and the fact $x\to x^{-1}$ is continuous. We care only for terms $n>\lfloor x\rfloor$ to make the above meaningful.

NOTE If you're acquainted with $\limsup$ and $\liminf$; the above can be put differently as $$\eqalign{ & \exp x \leqslant \lim \inf {\left( {1 + \frac{x}{n}} \right)^n} \cr & \exp x \geqslant \lim \sup {\left( {1 + \frac{x}{n}} \right)^n} \cr} $$ which means $$\lim \inf {\left( {1 + \frac{x}{n}} \right)^n} = \lim \sup {\left( {1 + \frac{x}{n}} \right)^n}$$ and proves the limit exists and is equal to $\exp x$.

$\endgroup$
  • $\begingroup$ What is $\text{exp}$? and $\text{sup}$? $\endgroup$ – Jeel Shah Apr 12 '13 at 2:47
  • $\begingroup$ What a job, Peter, +1. It remains $\lim (1+x/n)^n$... $\endgroup$ – Julien Apr 12 '13 at 2:50
  • 1
    $\begingroup$ @gekkostate $\exp$ stand for the "exponential" function. One doesn't really know beforehand that $\exp x=e^x$, but it is a very easy thing to see as I'm showing you. The "word" $\sup$ stands for the "supremum" of a set. In the case of real numbers, we define the supremum of a set $A\subseteq \Bbb R$ (if it exists) to be the (unique) number with the following properties: $\endgroup$ – Pedro Tamaroff Apr 12 '13 at 2:52
  • 1
    $\begingroup$ $(i)$ For any number $a\in A$, $a\leq x$. That is, $x$ is what we call an upper bound for the elements of $a$. As an example, $2$ is an upper bound for $[0,1)$, because for any $x\in[0,1)$, $x\leq 2$. $(ii)$ If $y$ is any other number with the above property $x\leq y$. That is, $x$ is the least of all the upper bounds. That is why we usually call $\sup$ the "least upper bound" of $A$, and write $\operatorname{lub}(A)$. ${}$ Question: What is $\sup [0,1)$? A tougher one: what is $\sup \{r\in \Bbb Q:r^2<2\}$? $\endgroup$ – Pedro Tamaroff Apr 12 '13 at 2:53
  • 1
    $\begingroup$ @gekkostate Let $e^h-1=u$. Then $u\to 0\iff e^h-1\to 0$. The limit then states, since $h=\log(u+1)$, $$\lim_{u\to 0}\frac{u}{\log(1+u)}=1$$ which we can prove using the definition of the logarithm I gave you. $\endgroup$ – Pedro Tamaroff Apr 13 '13 at 15:25
12
$\begingroup$

First prove that $\lim_{h\to 0}\frac{\ln(h+1)}{h}=1$. The switch of $\ln$, $\lim$ is possible because $f(x)=\ln x$, $x>0$ is continuous.

$$\lim_{h\to 0}\ln\left( (1+h)^{\frac{1}{h}} \right)=\ln\lim_{h\to 0}\left( (1+h)^{\frac{1}{h}} \right)=\ln e = 1$$

Now let $u=e^h-1$. We know $h\to 0\iff u\to 0$.

$$\lim_{h\to 0}\frac{e^h-1}{h}=\lim_{u\to 0}\frac{u}{\ln(u+1)}=1$$

$\endgroup$
4
$\begingroup$

If you have have a proof for $\displaystyle\frac{d\ln x}{dx}$, you can always prove $\displaystyle\frac{de^x}{dx}$ by taking $\displaystyle\frac{d\ln e^x}{dx}$ using chain rule.

First, we know: $$\frac{d\ln e^x}{dx} = \frac{dx}{dx} = 1$$

Second, using chain rule:

$$\frac{d\ln e^x}{dx} = \frac{1}{e^x}\frac{de^x}{dx} $$ Using the fact that we know both sides equal $1$, then:

$$\frac{1}{e^x}\frac{de^x}{dx} = 1 $$ Multiplying both sides by $e^x$ gives:

$$\frac{de^x}{dx}= e^x$$

$\endgroup$
2
$\begingroup$

To show the equivalence of the characterizations $e = \lim_{n \to \infty} (1 + \frac 1n)^n$ and $e^x = \frac d{dx} e^x$, assume the latter and prove that $\ln'(x) = \frac 1x$.

$$ 1 = \ln'(1) = \lim_{x \to 0} \frac {ln(1+x)-\ln(1)}x = \lim_{x \to 0} \frac 1x \ln(1+x) = \lim_{x \to 0} \ln\left((1+x)^{\frac 1x}\right) $$

Then, by the continuity of $\exp(x)$:

$$ e = e^1 = e^{\lim_{x \to 0} \ln\left((1+x)^{\frac 1x}\right)} = \lim_{x \to 0} e^{\ln\left((1+x)^{\frac 1x}\right)} = (1+x)^{\frac 1x} = \lim_{n \to \infty} \left(1+ \frac 1n\right)^n $$

Since the limit is well-defined, the characterizations must be equivalent.

$\endgroup$
  • $\begingroup$ This is a clever answer but how would you know ln(1)=0 if we're assuming the new definition of e? $\endgroup$ – Addem Jul 21 '16 at 3:31
0
$\begingroup$

HINT :for one definition of e $$e=\lim_{n\rightarrow\infty}(1+\frac1n)^n$$ and use the Binomial expansion

$\endgroup$
  • 1
    $\begingroup$ Could you elaborate? $\endgroup$ – Yatharth Agarwal Oct 12 '15 at 19:58
0
$\begingroup$

Just to prove this in a new way, using a specialized case:

$$\lim_{h \rightarrow 0} \frac{e^{ah}-1}{h}$$

where a is any number

Here's how I did it:

$$e= \lim_{n \to \infty} \Big(1 + \frac{1}{n} \Big)^{n}$$

Recognize that $1/n$ is just an infinitesimal, and $n$ is just a representation of infinity. By doing this, we can redefine $e$ in terms of $h$.

$$e= \lim_{h \to 0} \Big(1 + h \Big)^{\frac{1}{h}}$$

Plug this back in:

$$\lim_{h \rightarrow 0} \frac{e^{ah}-1}{h} = \lim_{h \rightarrow 0} \frac{((1+h)^{1/h})^h-1}{h} $$

$$= \lim_{h \rightarrow 0} \frac{((1+h)^{1/h})^{ah}-1}{h} $$

$$= \lim_{h \rightarrow 0} \frac{(1+h)^a-1}{h} $$

Binomial theorem:

$$= \lim_{h \rightarrow 0} \frac{\Big(\dbinom{a}{0}1^ah^0+\dbinom{a}{1}1^{a-1}h^1+ ... +\dbinom{a}{a-1}1^{1}h^{a-1}+\dbinom{a}{a}1^{0}h^a\Big)- 1}{h}$$

Getting rid of all of those pesky ones:

$$= \lim_{h \rightarrow 0} \frac{\Big(\dbinom{a}{0}h^0+\dbinom{a}{1}h^1+ ... +\dbinom{a}{a-1}h^{a-1}+\dbinom{a}{a}h^a\Big)-1}{h}$$

The term $\dbinom{a}{0}h^0$ is equal to one, allowing us to cancel the stuff at the end.

$$= \lim_{h \rightarrow 0} \frac{\Big(\dbinom{a}{1}h^1+ ... +\dbinom{a}{a-1}h^{a-1}+\dbinom{a}{a}h^a\Big)}{h}$$

We can now simplify the limit. $$= \lim_{h \rightarrow 0} \Big(\dbinom{a}{1}+ ... +\dbinom{a}{a-1}h^{a-2}+\dbinom{a}{a}h^{a-1}\Big)$$

We can now evaluate the limit. Everything other than the first term is multiplied by h and becomes 0, leaving us with

$$= \dbinom{a}{1} = a$$

Meaning $$\frac{e^{ah}-1}{h} = a$$. However, when a = 1, the first limit also equals 1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.