the value of $e$ and the method of getting it

Question

We define e to be a number which satisfies the following condition

$$\lim _{a \to 0} \frac{e^a-1}{a}=1. $$

How did we arrive to the following from above equation

$$e=\lim _{n \to \infty} \bigg(1+\frac{1}{n}\bigg)^n ? $$

so that we get the value of n

Answer 1

It's easier to start from the first equation and prove that it's the same as the second one.

From $$L := \lim _{a \to 0} \frac{e^a-1}{a}$$ set $e^a - 1 = t$, so that $a = \ln(1 + t)$: $$L = \lim_{t \to 0} \frac{t}{\ln(1 + t)} = \lim_{t \to 0} \left(\frac{\ln(1 + t)}{t}\right)^{-1} = \lim_{t \to 0} \left(\ln\left[\left(1 + t\right)^{1/t}\right]\right)^{-1} = \left(\ln\left[\lim_{t \to 0} \left(1 + t\right)^{1/t}\right]\right)^{-1}$$

The last step is legit because both $1/x$ and $\ln x$ are continuous functions. Now, recalling that $L = 1$ (from your first equation) we have that $$\begin{align} \ln\left[\lim_{t \to 0}\left(1 + t\right)^{1/t}\right] &= 1\implies\\ \lim_{t \to 0}\left(1 + t\right)^{1/t} &= e \end{align}$$

Or, equivalently, with one last change of variable $$e=\lim _{x \to +\infty} \left(1+\frac1x\right)^x.$$

Answer 2

The approach you mention is difficult, but possible. I have presented it in my blog post. The main steps are as follows:

1) Define $a^{b}$ (without using any logs or $e$) rigorously for $a > 0$ and any real $b$.

2) Show that $\lim_{a \to 0}\dfrac{x^{a} - 1}{a} = f(x)$ exists for all $x > 0$ and hence defines a function of $x$. This function is denoted by $\log x$ and $e$ is then a number such that $\log e = 1$.

3) With $\log x$ defined above we have the following properties: $\log(xy) = \log x + \log y, \log 1 = 0, \log(1/x) = -\log x, \log(x^{y}) = y\log x$.

4) $\lim_{x \to 0}\dfrac{\log(1 + x)}{x} = 1$

5) Putting $x = 1/n$ in above limit and get $\lim_{n \to \infty}\log\left(1 + \dfrac{1}{n}\right)^{n} = 1$ and noting that $\log e = 1$ we get $\lim_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^{n} = e$.

Proof of 1), 2) and 4) is hard but not too hard.