Explicit expression for $1/e$ as a limit

Question

I am following an elementary math book: What is Mathematics and currently referring to infinite series representation of the exponent.

In deriving the explicit formula for $e$ and $1/e$, the author exploits the derivative of the log function (see attached screenshots).

I am clear till the point $e^z$ is defined as $\lim_{n\to\infty}(1+z/n)^n$. After this, the author substitutes $z$ as $1$ and $-1$ to get expansions for $e$ and $1/e$.

Here, I'm not able to understand why its okay to take $z$ as $-1$

The way I see it, $z$ equals $1/x$ (screenshot 2) and $1/x$ cannot be negative because it's the derivative of the log (monotonically increasing function)

Additional notes:

The author has defined the log function as integral(1/u)du from 1 to x. (Screenshot 3)

He then proceeds to define the exponent as an inverse operation to this 'newly discovered' log function (allowed because in the range defined (0,infinity), log is a monotonically increasing continuous function)

And only then proceeds with the argument above.

Hence, I wonder how could we take z = -1. Wouldn't it imply 1/x = -1 i.e x = -1 which is outside the defined domain for logx?

If we rather take the function as log|x|, that opens the question, how do we define log|x| as an integral? (similar to the author's approach)

How do we define the inverse of log|x| (not 1-1)?

What am I missing here?

Answer 1

For any $z$ you can define $E(z)=\lim_{n\to\infty}\left(1+\frac zn\right)^n$. The cited argument shows that $E(z)=\exp(z)$ for positive $z$, where $\exp$ is the inverse function of $\ln$. In the cited pages it is not shown that $\exp$ is an exponential function, so that writing $\exp(z)=e^z$ is not really justified. For that one would need to show that the functional equation $E(x)E(y)=E(x+y)$ is satisfied, and that $E$ is a continuous function (some rather weak formulations are possible).

For any real number $z$ you get, for sufficiently large $n$, $$ \left(1+\frac zn\right)^n\left(1-\frac zn\right)^n = \left(1-\frac{z^2}{n^2}\right)^n\in\left[1-\frac{z^2}n,\,1\right] $$ the last using the Bernoulli inequality $(1+a)^n\ge 1+an$ for $a\ge -1$. From this follows $$ E(z)E(-z)=1\implies E(-z)=E(z)^{-1}. $$

This allows to extend the identification $E(z)=\exp(z)=e^z$ also to negative $z$.

Answer 2

Note that $\log\lvert x\rvert$ is differentiable in $\mathbb R\setminus\{0\}$ and if you differentiate it, you get $\frac1x$. So, yes, you can take negative values for $x$.

Answer 3

Notice that

$$1-\frac1n=\frac1{\dfrac n{n-1}}=\frac1{1+\dfrac1{n-1}}.$$

Then raising to the power $n$ or $n-1$ makes no difference (the factor tends to $1$) and

$$f(-1,n)=\frac1{f(1,n)}$$ in the limit.