*Actually* calculating 2.718…

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

Now, this is arguably the most popular definition of the infamous number e. And you also might have someone, most probably your teacher, give you this definition without really offering a proof or an explanation. Oh but wait, they also told you that e is the number which you get when an sum of money is continuously compounded. Looking at the limit, it might makes sense. However, if you are anything like me, you were wondering ok but why exactly is this number 2.718…?

One reason your teacher might not have given you an explanation is because what I am about to show you does require some notion of continuity and calculus. Another reason is that your teacher is simply ignorant… Let’s get started:

\displaystyle \lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x

We’re going to set this equal to a variable, \displaystyle y in this case, so we have an equation to work with:

\displaystyle y=\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x

Taking the natural log of both sides:

\displaystyle \ln(y)=\ln\left(\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x\right)

Ideally, what we want to do is to be able to interchange the limit and the function composition. So we want to know if

\displaystyle \lim_{x\to\infty}\left(\ln\left(1+\frac{1}{x}\right)^x\right)=\ln\left(\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x\right)

It turns out that to be able to do that requires that \displaystyle \ln\left(1+\frac{1}{x}\right)^x is continuous as \displaystyle x\to\infty. And this is the case. Since

\displaystyle \lim_{x\to\infty}\left(\ln\left(1+\frac{1}{x}\right)^x\right)=\ln\left(\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x\right)

\displaystyle \ln(y)=\lim_{x\to\infty}\left(\ln\left(1+\frac{1}{x}\right)^x\right)

\displaystyle \ln(y)=\lim_{x\to\infty}\left(x * \ln\left(1+\frac{1}{x}\right)\right)

Let’s calculate this limit using L’Hôpital’s Rule by brute forcing the limit into an indeterminate form:

\displaystyle \lim_{x\to\infty}\left(x * \ln\left(1+\frac{1}{x}\right)\right)

\displaystyle \lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}

\displaystyle =\frac{0}{0}

Applying L’Hôpital’s gives

\displaystyle \lim_{x\to\infty}\frac{\frac{d}{dx}\ln\left(1+\frac{1}{x}\right)}{\frac{d}{dx}\frac{1}{x}}

\displaystyle \lim_{x\to\infty} \frac{ \frac{-x^{-1}}{1+\frac{1}{x}}}{-x^{-1}}

\displaystyle = \lim_{x\to\infty} \frac{1}{1+\frac{1}{x}}

\displaystyle =1

Going back to our previous equations:

\displaystyle \ln(y)=\lim_{x\to\infty}\left(\ln\left(1+\frac{1}{x}\right)^x\right)

\displaystyle \ln(y)=1

\displaystyle y=e^1=e

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

I will admit that this way of calculating e requires some previous concept of the natural log, and hence e itself. But since we don’t live in the 17th century, this derivation works fine for us. Anyways, these questions are typically left to the epistemologists, right? But aren’t mathematicians epistemologists?…

 

 

 

 

 

 

 

 

 

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