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I was recently thinking about how I could evaluate the famous limit of 'e' as I haven't ever seen a proof. I can't really find anything online so I've tried to evaluate the limit myself. And I was also thinking it would be nonsensical to use L'Hopital's rule, am I right?

So I did the following:

$$\lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^{n} = \exp \left(\lim_{n \to \infty} n \cdot \ln \left(1 + \frac{1}{n} \right) \right)$$

$$=\lim_{n \to \infty} \exp \left( n \cdot \left( \frac{1}{n} - \frac{(1/n)^{2}}{2} + \cdots \right) \right)$$

$$= \lim_{n \to \infty} \exp \left( 1 - \frac{(1/n)}{2} + \cdots \right)$$

$$= e$$

I am not sure, is my logic correct or does it create circularity by taking logarithms and assuming $$f(x) = \exp \left( \ln f(x) \right)$$ ?

Cameron Buie
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user78416
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    This is indeed a standard technique for handling limits of this form. Similarly to compute e.g. $$\lim_{x \to 0} x^x$$ one takes logs, deals with $x \ln x$ via L'Hospital's Rule or series expansion, and exponentiates the result. – gt6989b Jun 11 '13 at 16:08
  • I see. I was just afraid that the logic may be a bit circular by assuming the natural log which leads to L'Hopital's rule/ series expansions of ln – user78416 Jun 11 '13 at 16:11
  • Your concern seems rooted in whatever the definition of the functions $\exp$, $\ln$, and the number $e$ are. There are many different definitions each of these things, all of which can be viewed as consequences of any of the others. I suggest you solidify what your definitions are of these three things, and then see if you can establish (based on their definitions) the features that you use: that $\ln$ and $\exp$ are inverses, the power series for $\ln$, and that $e=\exp(1)$. Some of these features may be true by definition. – 2'5 9'2 Jun 11 '13 at 16:21
  • I've proved the 'e' method in my answer here: http://math.stackexchange.com/questions/447312/evaluate-lim-x-to0-left-frac-sin-xx-right6-x2/447374#447374 :) – mikhailcazi Oct 31 '13 at 12:21

4 Answers4

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Taking logs is a straightforward approach. I prefer to use a first order expansion based on the fact that $\ln$ is differentiable, rather than a series expansion, but they amount to the same thing here.

If $\phi(x) = \ln (x+1)$, then $\phi(x) = 0 + 1 \cdot x + o(x)$, where $\frac{o(x))}{x} \to 0$, as $x \downarrow 0$. Hence $\lim_{x \downarrow 0} \frac{\phi(x)}{x} = 1$. It follows that $\lim_{n \to \infty} n \ln(1+ \frac{1}{n}) = 1$.

Since $x \mapsto e^x$ is continuous, we have $\lim_{n \to \infty} e^{n \ln(1+ \frac{1}{n})} = \lim_{n \to \infty} (1+\frac{1}{n})^n = e^1 = e$.

copper.hat
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  • Can you use $\ln$ and the continuity of $e^x$ to try to define $e$ though? – N. S. Jun 11 '13 at 16:18
  • @N.S.: My preferred approach is to define $\exp$ by series, and $\ln$ as the inverse (using the inverse function theorem as appropriate). (Is this what you were asking?) – copper.hat Jun 11 '13 at 16:21
  • @N.S.: Actually, I think a nicer approach is to define $\exp$ as the solution of $\dot{x} = x$ with initial condition $x(0) = 1$ (and then use the IFT, etc.) but this involves a little more analytic machinery. – copper.hat Jun 11 '13 at 16:28
  • Well it really boils down to the definition of $e^x$. Anyhow, the point is that if you define exp or ln that way, there is no reason to define then $e$ as the limit of the above sequence. And, while I agree that some definition make things easier to use, there are really using higher artillery, and in some sense they are backwards..... The "standard" proof of $e$ being the limit of that sequence uses very little analysis, as opposite to introducing Power/Taylor series.... – N. S. Jun 11 '13 at 16:34
  • @N.S.: How is $e$ defined in the standard proof? – copper.hat Jun 11 '13 at 16:41
  • As far as I know, $e$ was defined historically as the limit of the sequence in the question. You can prove then from that limit that $(e^x)'=e^x$, and obtain the other easy characterizations of $e^x$. – N. S. Jun 11 '13 at 16:45
  • @N.S.: Thanks! That refreshed some really old secondary school memories. – copper.hat Jun 11 '13 at 16:53
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To avoid circularity, it depends on what you define how, e.g. some authors define the number $e$ by $\lim(1+\frac1n)^n$. Of course, then they need to show that $\exp(1)=e$.

The prettiest (in my opinion) introduction of $e,\exp,\ln$ is by showing that the space of solutions of the differential equation $f'(x)=f(x)$ is onedimensional. Then let $\exp$ be the solution with $\exp(0)=1$, immediately obtain $\exp(x+y)=\exp(x)\exp(y)$ (by uniqueness of solution), $\exp(nx)=\exp(x)^n$, $\exp(x)\ne0$, etc. Then we define $e:=\exp(1)$ (because that allows us to use the suggestive notation $e^x$ for $\exp(x)$).

Now in that setup the limit can be obtained without even digging deeper into l'Hospital and $\ln$ (apart from its local existence): Since $\exp'(0)=1\ne0$, the function does have an inverse near $x=0$, hence for $n$ big enough, there is $x_n\approx 0$ such that $\exp(x_n)=1+\frac1n$ (you can even avoid invoking the inverse/implicit function theorem and yous get along with the intermediate value theorem). Since $1=\exp'(0)=\lim_{n\to\infty}\frac{\exp(x_n)-\exp(0)}{x_n}$, we get $nx_n\to 1$ and hence $(1+\frac1n)^n=\exp(nx_n)\to \exp(1)=e$.

Hagen von Eitzen
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There are several ways to approach this question.

In this answer, it is shown that $$ \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\sum_{k=0}^\infty\frac{x^n}{n!} $$ which gives the usual value for $e^x$, in particular, $e=e^1$.

Note that $$ a_n=\left(1+\frac1n\right)^n\lt\left(1+\frac1n\right)^{n+1}=b_n $$ In this answer, it is shown that $a_n$ is an increasing sequence and $b_n$ is a decreasing sequence. Thus, $a_n$ is an increasing sequence bounded above by $a_n\lt b_n\lt b_1=4$ and $b_n$ is a decreasing sequence bounded below by $2=a_1\lt a_n\lt b_n$. Thus, both sequences converge.

Since $$ \lim_{n\to\infty}\frac{b_n}{a_n}=\lim_{n\to\infty}1+\frac1n=1 $$ the limits of both sequences are the same. Call that limit $e$, then we have $$ 2=a_1\lt\lim_{n\to\infty}a_n=e=\lim_{n\to\infty}b_n\lt b_1=4 $$ By computing $a_n$ and $b_n$ for large $n$, we can bound the value of $e$ as closely as we want.

Community
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robjohn
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Here is the "standard" proof:

Let $a_{n}= \left(1 + \frac{1}{n} \right)^{n}= \frac{(n+1)^n}{n^n}$.

Then

$$\frac{a_{n+1}}{a_n}=\frac{(n+2)^{n+1}}{(n+1)^{n+1}}\frac{n^n}{(n+1)^n}=\frac{(n+2)^{n+1}n^{n+1}}{(n+1)^{2n+2}}\frac{n+1}{n}$$

$$=\left( \frac{n^2+2n}{(n+1)^{2}}\right)^{n+1}\frac{n+1}{n}=\left( 1-\frac{1}{(n+1)^{2}}\right)^{n+1}\frac{n+1}{n}$$

Now, by Bernoulli

$$\left( 1-\frac{1}{(n+1)^{2}}\right)^{n+1}\geq 1-\frac{n+1}{(n+1)^{2}}=\frac{n}{n+1}$$

Thus

$$\frac{a_{n+1}}{a_n} \geq \frac{n}{n+1}\frac{n+1}{n}=1 \,.$$

This shows that $a_n$ is increasing.

Now, let

Let $b_{n}= \left(1 + \frac{1}{n} \right)^{n+1}= \frac{(n+1)^{n+1}}{n^{n+1}}$.

Then

$$\frac{b_n}{b_{n+1}}=\frac{(n+1)^{n+1}}{n^{n+1}}\frac{(n+1)^{n+2}}{(n+2)^{n+2}}=\left( \frac{(n+1)^2}{n(n+2)} \right)^{n+2}\frac{n}{n+1}$$

$$\geq \left(1+ \frac{n+2}{n(n+2)} \right)\frac{n}{n+1}=1$$

Thus $b_n$ is decreasing.

As $a_n \leq b_n$ it follows that $a_n$ is increasing and bounded from above by $b_1$..

Now, if we denote the limit of this sequence by $e$ (which is the historic definition), it can be proven from

$$\left(1 + \frac{1}{n} \right)^{n}=\left(1 + \frac{1}{n} \right)^{n+1} =e$$ that $(e^x)'=e^x$.

N. S.
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  • I think the first proof I saw involved another approach to the upper bound as $(1+\frac{1}{n})^n = \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} \le \sum_{k=0}^n \frac{1}{k!} < \sum_{k=0}^\infty \frac{1}{k!}$, but this involved the ratio test. – copper.hat Jun 11 '13 at 17:05