Is there a way to remember the limit definition for $e$?

Question

I occasionally see proofs where the limit definition for $e$ pops up and I don't recognize it for some reason, every time!

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

For one thing I don't know why it's defined this way or why this particular number is so important. I know the derivative of $e^x$ is still $e^x$ so there's probably some relation in there (except when I take the derivative of $e$ in Wolfram Alpha it gives me a huge messy log expression, not equal to the original), but how can I better understand what $e$ is so all these various relationships and proofs are more obvious to me?

For instance that limit expression I have no idea how to solve, because the inner part approaches $1$ but the exponent approaches infinity, but the answer is certainly not $1^\infty = 1$.

Why is it defined this way? How do we solve it? How do we use it?

Answer 1

$$\begin{aligned} \lim_{n \to \infty} \left( 1 + \frac{1}{n}\right)^n&=\lim_{n \to \infty}\exp\left(\ln\left( \left( 1 + \frac{1}{n}\right)^n\right)\right)=\exp\left(\lim_{n \to \infty}\ln\left( \left( 1 + \frac{1}{n}\right)^n\right)\right)=\\& =\exp\left(\lim_{n \to \infty}n\ln \left( 1 + \frac{1}{n}\right)\right)\to\small{\begin{bmatrix} &s=1/n&\\ &s\to0& \end{bmatrix}}\to\\ &\to\exp\left(\lim_{s\to 0}\frac{\ln\left( 1 + s\right)}{s}\right)=\exp(1)=e. \end{aligned}$$

Note: This could have also worked with $\lim_{n \to \infty}a^{\log_a\left( 1 + \frac{1}{n}\right)^n}$ with $a$ being a constant, but here Hôspital's Rule must be used. (Be careful when you differentiate $\log_a(f(x))$).

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Another approach: (requires Calculus II knowledge)

Using the Binomial Theorem

$$\left( 1 + \frac{1}{n}\right)^n=\sum_\limits{j=0}^\limits{n}{n \choose j}\frac{1}{n^j}=\sum_\limits{j=0}^\limits{n}\frac{n!}{(n-j)!n^j}\frac{1}{j!}$$ Note that $$\left(\frac{n-j+1}{n}\right)\leq\frac{n!}{(n-j)!n^j}\leq1$$ and by Bernoulli's inequality $$\left(\frac{n-j+1}{n}\right)^j=\left(1-\frac{j-1}{n}\right)^j\geq1-\frac{j(j-1)}{n}$$ whence $$\sum_\limits{j=0}^\limits{n}\frac{1}{j!}-\frac{1}{n}\sum_\limits{j=0}^\limits{n}\frac{1}{(j-2)!}\leq\left( 1 + \frac{1}{n}\right)^n\leq\sum_\limits{j=0}^\limits{n}\frac{1}{j!}$$ Since $\sum_\limits{j=0}^\limits{n}\frac{1}{j!}$ is convergent it follows that $\left( 1 + \frac{1}{n}\right)^n$ tends to the sum of that series, which is $e$.

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If you want to know more about this check out e: the Story of a Number by Eli Maor.

Answer 2

• Why is it defined this way?

The fundamental reason is that $(e^x)'=e^x$ and that makes it the most "natural" choice of the base. By contrast, $(a^x)'=\log a\,a^x$, and we don't like the $\log a$ factor.

• How do we solve it?

Strictly speaking you can't solve the limit $\lim_{n\to\infty}\left(1+\frac1n\right)^n$ because it is the definition of the constant $e$. As long as you have not defined $e$, you cannot express the value of the limit as it is a not-yet-defined number. (Some other answers here are circular arguments showing that $e=e$.)

[A similar situation is with $\lim_{n\to\infty}\left(\sum_{k=1}^\infty\frac1k-\log n\right)$, which tends to a "new" constant named $\gamma$.]

• How do we use it?

A convenient way to evaluate it is by turning the definition in the equivalent expression $\sum_{n=0}^\infty \frac1{n!}$, which is a very fast convergent series. This formula can be obtained by expanding $(1+\frac1n)^n$ by means of the binomial theorem and reworking.

It is also useful to know the formula below (easily obtained by the change of variable $nx\leftrightarrow m$):

$$e^x=\lim_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim_{m\to\infty}\left(1+\frac xm\right)^m.$$

• Is there a way to remember the limit definition ?

Maybe it helps to remember that it is a $1^\infty$ indeterminate form (hence some connection to exponentials), where $1$ is reached in the simple form $1+\frac1n$ and infinity as just $n$. That's about the simplest indeterminate form $1^\infty$ you can think of.

Answer 3

Well I don't know for a fact that this limit definition came from compound interest, but that's certainly how I was introduced to it. In one year at an interest rate of $r$ your gain is a factor of $(1 + r)$. Compounded twice a year at the same annual rate, it is $(1 + \frac{r}{2})^2$. Twelve times a year would be $(1 + \frac{r}{12})^{12}$. If you compounded interest continuously and your interest rate was 100% (quite the bank!) then you'd have $\lim_{n\rightarrow \infty} (1 + \frac{1}{n})^n$. To prove that this converges and is the base of the natural logarithm takes some work.

Answer 4

The more relevant limit is $$ e^x = \lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n $$ which comes from compound interest problems. This, I believe, was the origin of $e$, which is obviously the case $x = 1$.

I'm going to demonstrate that this is indeed $e^x$, which will also tie this together with several other definitions of the function.

First, we need to know this limit converges. We look at $$ \left(1+\frac{x}{n}\right)^n = \sum_{i=0}^n\frac{n!}{i!(n-i)!}\frac{x^i}{n^i} = \sum_{i=0}^n\left[\prod_{j=1}^i\left(1+\frac{j+1}{n}\right)\right]\frac{x^i}{i!} $$ For every $i$, that product is a finite number of factors that converge to $1$ as $n\rightarrow\infty$, so it also converges to $1$. Thus, $$ \lim_{n\rightarrow\infty}\left(1 + \frac{x}{n}\right)^n = \sum_{i=0}^\infty\frac{x^i}{i!} $$ which can be shown to converge by your favorite series convergence test. Note that this is, in fact, the power series for $e^x$.

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Next we show that this is $\exp(x)$, the function whose derivative is equal to itself and satisfies $\exp(0) = 1$. That second part is self-evident, while a quick application of the chain rule shows $$ \frac{d}{dx}\left[\left(1+\frac{x}{n}\right)^n\right] = n\left(1+\frac{x}{n}\right)^{n-1}\frac{1}{n} = \left(1 + \frac{x}{n}\right)^{n-1} = \frac{1}{1+x/n}\cdot\left(1+\frac{x}{n}\right)^n. $$ Using that in our limit definition gives $$ \frac{d}{dx}\lim_{n\rightarrow\infty}\left[\left(1+\frac{x}{n}\right)^n\right] = \left[\lim_{n\rightarrow\infty}\frac{1}{1+x/n}\right]\left[\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n\right] = \lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n $$ as desired. Note that termwise differentiation of $\sum x^i/i!$ will also give you back the same function, so these both represent $\exp(x)$. That is, $$ \lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n = \sum_{i=0}^\infty \frac{x^i}{i!} = \exp(x) $$

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Lastly, we show that $\exp(x + y) = \exp(x)\exp(y)$. This can be found by noting that $(d/dx)[\exp(x + y)] = \exp(x+y)$, so $\exp(x+y) = C\exp(x)$ for some constant $C$, which pretty clearly has to be $\exp(y)$. So $\exp(x+y) = \exp(x)\exp(y)$. This can be extended to show that $\exp(x) = \exp(1)^x$ for all rational $x$, at which point continuity ensures it's true for all $x$. Letting $e = \exp(1)$ gives $$ \lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n = \sum_{i=0}^\infty \frac{x^i}{i!} = \exp(x) = e^x $$ which gives us as a special case $$ \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n = \sum_{i=0}^\infty \frac{1}{i!} = \exp(1) = e $$

Answer 5

The reason $e^x$ is called the natural exponential function and $\ln x= \log_e x$ is called the natural logarithm is that they occur naturally as you develop calculus. For example, you can prove from definition that for any function $f(x)=a^x$ where $a>0$, the derivative $f'(x)$ is given by $$ f'(x)=a^x\cdot\lim_{h\to 0}{\frac{a^h-1}{h}}=a^x\cdot C_a, $$ where $C_a$ is some constant depending on $a$, but independent of $x$. It is also pretty easy to see that if $0<a<b$, then $C_a<C_b$, and that $C_1=0$.

Naturally, you would prefer not to have to lug this constant around in all your calculations involving the exponential function, nor to have an extra factor appear every time you differentiate such a function. In other words, you would like to find some constant $e$ for which $C_e=1$.

After some more calculations involving the chain rule and the inverse function, it can be shown that $(\ln x)'=\frac{1}{x}$, and since by the change-of-base formula we have $\log_a x=\frac{\ln x}{\ln a}$, it turns out that $(\log_a x)'=\frac{1}{x\ln a}$, and from that using the derivative of the inverse function again, we conclude that $(a^x)'=a^x\ln a$. And thus we need the number $e$ pretty much everywhere in calculus.

At the same time, thinking about the continuously compounded interest problem, one can show that all such problems can be reduced to \$1 continuously compounded at 100% nominal annual interest. The amount you get from this is $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$.

It can be proved using the AM-GM inequality and the forward-backward induction that the sequence $a_n=\left(1+\frac{1}{n}\right)^n$ is increasing, while the sequence $b_n=\left(1+\frac{1}{n}\right)^{n+1}$ is decreasing, which means (since $b_n=\left(1+\frac{1}{n}\right)a_n>a_n$ for every $n$) that both sequences are bounded by $a_1=2$ from below and $b_1=4$ from above, and thus both sequences converge, and to the same limit. It can be shown (see one of the other answers here) that this limit is the same number $e$ as the one we naturally needed for differentiation of exponential functions.

So, "why is it defined this way?" Because we need it in so many places, not for any cruel or unusual purposes. Both $e$ and $\pi$ are complicated numbers, but it's a trade-off for all the nice properties they have.