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?

  • $\begingroup$ But you're familiar with $$e^x=\lim\limits_{h\to\infty}\left(1+\frac{x}{h}\right)^h$$ aren't you? $\endgroup$ – J. M. is a poor mathematician Dec 3 '11 at 6:36
  • $\begingroup$ @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? $\endgroup$ – EuYu Dec 3 '11 at 6:48
  • $\begingroup$ I would go with using the binomial expansion, you see... $\endgroup$ – J. M. is a poor mathematician Dec 3 '11 at 6:59

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...


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.

  • $\begingroup$ 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 $f'(x) = e^x \ln(e)$? $\endgroup$ – EuYu Dec 19 '11 at 1:10
  • $\begingroup$ For that simplification click "show steps" here: wolframalpha.com/input/?i=lim+h-%3E0+%28e%5Eh+-1%29%2Fh 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. $\endgroup$ – Samuel Reid Dec 19 '11 at 3:01
  • $\begingroup$ 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. $\endgroup$ – EuYu Dec 19 '11 at 4:09
  • $\begingroup$ 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. $\endgroup$ – Samuel Reid Dec 19 '11 at 5:43

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