Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A common definition of $e$ is given as $$e = \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}$$ which can be proven to be equivalent to $$e=\lim_{h\rightarrow 0}\ \left(1+h\right)^{\frac{1}{h}}$$ The most practical use of $e$ in elementary calculus is however given as $$1=\lim_{h\rightarrow 0}\frac{e^h - 1}{h}$$ which is used as a statement the slope of $e^x$ at $x=0$ is $1$ allowing one to prove that $\frac{d}{dx}e^x=e^x$. It appears trivial to prove that the two limits given above are equivalent, but I cannot seem to make any progress without making some illegal limit operations. I suspect the problem is deeper than it appears (I suspect the trouble is that although we have defined $e$, we have not actually said anything about what $e^x$ is). How does one rigorously proceed from the given definition of $e$ to the slope limit?

share|cite|improve this question
But you're familiar with $$e^x=\lim\limits_{h\to\infty}\left(1+\frac{x}{h}\right)^h$$ aren't you? – J. M. Dec 3 '11 at 6:36
@J.M. Yes I am. Admittedly I was taking a shot in the dark about what the actual difficulty of the problem is. Does this help in anyway for actually proving the equivalence? – EuYu Dec 3 '11 at 6:48
I would go with using the binomial expansion, you see... – J. M. Dec 3 '11 at 6:59
up vote 1 down vote accepted

It is easy to go from the first limit to the second and back by using the substitution

$$z=\ln(1+h) \,,$$

Note that this is equivalent to $h= e^{z}-1$.

and the continuity of the $\ln()$.

$$e=\lim_{h\rightarrow 0}\ \left(1+h\right)^{\frac{1}{h}} \Leftrightarrow $$ $$1=\lim_{h\rightarrow 0}\ {\frac{\ln \left(1+h\right)}{h}} \Leftrightarrow$$ $$1=\lim_{z\rightarrow 0}\ {\frac{z}{e^z-1}} $$

Writing down a formal proof is easy now, you just have to be carefull filling it the details. You need to use the continuity of the logarithm function, the fact that $z=\ln(1+h)$ is a bijection from a neighborhood of zero to a neighborhood of zero; and most importantly that for each of the two implications $e$ and hence $\ln$ are defined in a different way. Thus, you cannot really work with if and only if, because then $\ln$ makes no sense. But you can do each implication separately easely...

share|cite|improve this answer

I'm not sure if you have multiple questions here but I'll try to answer all of them. For deriving: $$1=\lim_{h\rightarrow 0}\frac{e^h - 1}{h}$$

We start with the definition of a derivative:

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

If we let $f(x)=a^{x}$, where $a \in \mathbb{R}$ then,

$$f'(x) = \lim_{h \rightarrow 0} \frac{a^{x+h} - a^{x}}{h}$$ $$f'(x) =\lim_{h \rightarrow 0} \frac{a^{x}(a^{h} - 1)}{h}$$ $$f'(x) =a^{x} \left( \lim_{h \rightarrow 0} \frac{a^{h} - 1}{h} \right)$$ If we let $a=e$, since $e \in \mathbb{R}$ we have that $$f'(x) = e^{x} \left( \lim_{h \rightarrow 0} \frac{e^{h} -1}{h} \right)$$ Using l'hopital's rule and that $e^{h}$ is continuous at $h=0$ we can say that, $$f'(x)= e^{x} \ln(e)$$ So, if we evaluate it at $x=0$ we have, $$f'(0)=e^{0}=1$$ So that is how you obtain it one way.

Another way would go as follows: Let $f(x)=\ln(x)$, then $f'(x) = \frac{1}{x}$, so $f'(1)=1$ and then we say, $$1= \lim_{h \rightarrow 0} \frac{f(1+h) - f(1)}{h}$$ Because I like x's $$1 = \lim_{x \rightarrow 0} \frac{f(1+x) - f(1)}{x}$$ $$1 = \lim_{x \rightarrow 0} \frac{\ln(1+x) - \ln(1)}{x}$$ $$1 = \lim_{x \rightarrow 0} \frac{1}{x} \ln(1+x)$$ $$1 = \lim_{x \rightarrow 0} \ln(1+x)^{\frac{1}{x}}$$ Raise both sides by $e$ and we get, $$e^{1} = e^\left({\lim_{x \rightarrow 0} \ln(1+x)^{\frac{1}{x}}}\right)$$ $$e = \lim_{x \rightarrow 0} \text{ }e^\left({\ln(1+x)^{\frac{1}{x}}}\right)$$ $$e = \lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}}$$ And there you have your second formulation. For the first one, just let $x = \frac{1}{n}$ and you receive, $$e = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^{n}$$

Hopefully that helps! Let me know if I made a mistake somewhere.

share|cite|improve this answer
I appreciate the answer. However I am primarily interested in how the limit in question in obtained from the limit definition of $e$. Nevertheless your response is quite informative so I have given it +1. One question: How did you go from $f'(x) = e^x\left(\lim_{h\rightarrow 0}\frac{e^h - 1}{h}\right)$ to $f'(x) = e^x \ln(e)$? – EuYu Dec 19 '11 at 1:10
For that simplification click "show steps" here: As for how you obtain the last limit definition for e from the first one, take a look at the answer given by N.S., although he didn't exactly give a full explanation. – Samuel Reid Dec 19 '11 at 3:01
The method given requires the knowledge that $\frac{d}{dx}e^x = e^x$. Doesn't that make the limit rather trivial? It also seems too circular for my liking. – EuYu Dec 19 '11 at 4:09
Where does it circular? You start with the fact that $\frac{d}{dx} \ln(x) = \frac{1}{x}$ and from there you derive the definition of e. – Samuel Reid Dec 19 '11 at 5:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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