Proof? of the limit$ \lim_{n \to \infty}(1+\frac{1}{n})^{n} = e $

Question

we do know the formula for e: $$ \lim_{n \to \infty}(1+\frac{1}{n})^{n} = e $$ I was hoping to prove it using LH. Assume we don't know the value $$\lim_{n \to \infty}(1+\frac{1}{n})^{n} = x $$ Then we can use logarithms to solve this and since $$ \ln(x) $$ is continuous everywhere we can write $$\lim_{n \to \infty}n \ln(1+\frac{1}{n}) = \ln x $$ apply LH: $$\lim_{n \to \infty}\frac{\ln(1+\frac{1}{n})}{\frac{1}{n}} = \ln x $$ $$\lim_{n \to \infty}\frac{\frac{1}{(1+\frac{1}{n})}\frac{-1}{n^2}}{\frac{-1}{n^2}} = \ln x $$ this simplifies to $$\lim_{n \to \infty}\frac{1}{1+\frac{1}{n}} = \ln x $$ which we can evaluate to $$ \frac {1}{1+0} = 1 = \ln x$$ Therefore we come at $$ \ln x = 1 $$ $$x=e$$ What I'm wondering about is whether i am actually able to use the natural logarithm because i assume i don't know what 'e' is. Is this actually validor i have to take the limit purely as a definition of the euler number?

Answer 1

A person can define $\ln(x)$ and prove its derivative is $1/x$, etc, without knowing anything about $(1+1/n)^n$:

For $x>0$ define $$\ln(x)=\int_1^x\frac{dt}t.$$

It follows that $\ln'(x)=1/x$. Hence $\ln$ is strictly increasing on $(0,\infty)$. And it's clear that $$\lim_{x\to\infty}\ln(x)=\int_0^1\frac{dt}t=\infty.$$Similarly $\lim_{x\to0}\ln(x)=-\infty$.

So $\ln$ is a strictly increasing differentiable bijection from $(0, \infty)$ onto $\mathbb R$. So it has a strictly increasing differentiable inverse. Define $\exp:\mathbb R\to(0,\infty)$ to be the inverse of $\ln$. Define $e=\exp(1)$.

Note that $$\ln(xy)=\int_1^x\frac{dt}t+\int_x^{xy}\frac{dt}t=\int_1^x\frac{dt}t+\int_1^y\frac{dt}t=\ln(x)+\ln(y),$$by a simple substitution in the last integral. Hence $\ln(x^n)=n\ln(x)$ by induction; hence $\exp(nx)=\exp(x)^n$.

Now you don't even need L'Hopital, just the definition of the derivative shows that $$\lim_{n\to\infty}n\ln(1+1/n)=\lim_{n\to\infty}\frac{\ln(1+1/n)-\ln(1)}{1/n}=\ln'(1)=1/1=1.$$ Since $\exp$ is the inverse of $\ln$ and $\exp$ is continuous this shows that $$\lim(1+1/n)^n=\lim\exp(n\ln(1+1/n))=\exp(1)=e$$.

Answer 2

All of the objects involved here ($e, \ln x$, etc.) have several equivalent definitions and it takes some interesting work to prove that they're all equivalent; until you do questions like "how do I prove X" are ambiguous because they might be easy using some definitions and hard using others (before you've proven that they're all equivalent).

For example, you can define $\ln x$ without knowing what $e$ is, by defining it as the integral

$$\ln x = \int_1^x \frac{dt}{t}.$$

Using this integral you can prove the fundamental properties that $\ln (ab) = \ln a + \ln b$ and also that $\ln (1 + x) \approx x$ for $x$ small, which is enough to run your proof. Now you can define $e^x$ as the inverse of $\ln x$, if you want.

---

My preferred way to prove all of these equivalences is to start from the definition that $e^x$ is the unique function $f(x)$ satisfying $f(0) = 1$ and $f'(x) = f(x)$. From here, you can show that

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

by showing that this limit defines a function $g(x)$ which also satisfies $g(0) = 1$ and $g'(x) = x$; hence it must be equal to $e^x$, and now you can plug in $x = 1$. This identity can be interpreted as using Euler's method with smaller and smaller step sizes to approximate the solution to $f'(x) = f(x)$.

Generally speaking it's much nicer to prove things about $e^x$ rather than $e$, and arguably it is by far the more fundamental object.

Answer 3

I supplied an elementary proof (not my own) that $\lim_{n \to \infty} (1+1/n)^n$ exists here:

What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?

$e$ is then the name of that limit.

Once you know that the limit exists you can then do a variety of things to show that it is the same as your definition of $e$.