Understanding a particular proof of the derivative of $e^x$

Question

There are probably more efficient and easier proofs for same thing. This is proof I have to study for my exam.

Theorem: $(e^x)'=e^x$

Proof: For $x\ge 0$ we have defined $f_0(x)=\lim_{n\to +\infty}\left(1+\frac{x}{n}\right)^n.$ Exponential function is now defined by $$ f(x)=e^x= \begin{cases} f_0(x), & x\ge0 \\[4pt] \dfrac{1}{f_0(-x)}, & x\lt0 \end{cases}$$

For $x,c\gt0$ is

$$\frac{\left(1+\frac{x}{n}\right)^n-\left(1+\frac{c}{n}\right)^n}{x-c}=\frac{1}{n}\left[\left(1+\frac{x}{n}\right)^{n-1}+\left(1+\frac{x}{n}\right)^{n-2}\left(1+\frac{c}{n}\right)+\cdots+\left(1+\frac{x}{n}\right)\left(1+\frac{c}{n}\right)^{n-2}+\left(1+\frac{c}{n}\right)^{n-1}\right]$$

We have

$$\left(1+\frac{\min{(x,c)}}{n}\right)^{n-1} \le \frac{\left(1+\frac{x}{n}\right)^n-\left(1+\frac{c}{n}\right)^n}{x-c}\le \left(1+\frac{\max{(x,c)}}{n}\right)^{n-1}$$

For $n \rightarrow +\infty$ we have $$f_0(\min{(x,c))} \le \frac{f_0(x)-f_0(c)}{x-c} \le f_0(\max{(x,c))}$$

Functions $x \rightarrow \max{(x,c)}$ and $x \rightarrow \min{(x,c)}$ are continuous so for $x \rightarrow c$ is $$f_0^{'}(c) = \lim_{x\to c} \frac{f_0(x)-f_0(c)}{x-c} = f_0(c)$$

Now , for $ x\lt 0$ we have $f(x)=\frac{1}{f_0(-x)}$ , so$$ f'(x)=-\frac{f_0^{'}(-x)(-1)}{[f_0(-x)]^2}=\frac{f_0(-x)}{[f_0(-x)]^2}=\frac{1}{f_0(-x)}=f(x)$$

In case $c=0$ we have to separately look at left and right limit of $\frac{f(x)-1}{x}$. For $x\gt 0$ we have $1\le \frac{f_0(x)-1}{x} \le f_0(x)$, because of continuity of $f_0$ in $0$ and $f_0(0) = 1$ we have $$ \lim_{x\to 0+} \frac{f(x)-1}{x}=1=f(0) $$

For $x<0$ we have

$$\lim_{x\to 0-} \frac{f(x)-1}{x} = \lim_{x\to 0-} \frac{\frac{1}{f_0(-x)}-1}{x} = \lim_{-x\to 0+}\frac{f_0(-x)-1}{-x} \frac{1}{\lim_{-x\to 0+}f_0(-x)}=1=f(0)$$

Thus, $f'(0)=f(0)$. So, for $\forall x \in \mathbb{R}$ is $(e^x)'=e^x$.

---

I don't understand couple of things about this proof:

1. How does $(1+\frac{\min{(x,c)}}{n})^{n-1}$ go to $f_0(\min{x,c})$. We haven't got exponent $n$, but $n-1$?

2. Why is this valid: $ 1\le \frac{f_0(x)-1}{x} \le f_0(x)$ for $x\gt0$

3. Why is this valid : $\lim_{x\to 0-} \frac{\frac{1}{f_0(-x)}-1}{x} = \lim_{-x\to 0+}\frac{f_0(-x)-1}{-x} \frac{1}{\lim_{-x\to 0+}f_0(-x)}=1$

And one personal question, what do you think about this proof, and did your teacher in college requested something similar for you to know for exam?

Answer 1

2. is still incomplete, I'm still thinking about it. Hope the rest helps though.

1. Hint:

$$\lim_{n\to\infty}\left(1+\frac xn\right)^{n-1}=\lim_{k\to\infty}\left(1+\frac xk\right)\lim_{n\to\infty}\left(1+\frac xn\right)^{n-1}$$

3.

$$\lim_{x\to0^-}\frac{\frac1{f_0(-x)}-1}x=\lim_{-x\to0^+}\frac{f_0(-x)-1}{-x}\frac1{f_0(-x)}=1$$

$$\lim_{-x\to0^+}\frac{f_0(-x)-1}{-x}\frac1{f_0(-x)}=\lim_{-x\to0^+}\frac{\frac{f_0(-x)-1}{f_0(-x)}}{-x}=\lim_{-x\to0^+}\frac{1-f_0(-x)}{-x}=\lim_{-x\to0^+}\frac{f_0(-x)-1}{x}$$

It seems you are given $f_0(x)=\frac1{f_0(-x)}$ to conclude the left side equality.

See 2. to apply squeeze theorem on $f(x)$ to evaluate the limit to $1$.

Answer 2

I think both the definition of $e^{x}$ and the proof for its derivative are handled in a very roundabout way. Two definitions (one for $x \geq 0$ and one for $x < 0$) are an overkill when the same limit works for all real $x$. Moreover the inequality $1 \leq \dfrac{f_{0}(x) - 1}{x} \leq f_{0}(x)$ is not easy to establish and definitely not obvious. Your third question is made simple by writing $y$ in place of $-x$ and using $y \to 0^{+}$ instead of writing $-x \to 0^{+}$. This is a standard practice and we call it the method of substitution and here we have put $ x= -y$ to evaluate the limit.

A much better way is to proceed in the following manner. First show that for any given real number $x$ the limit $$\lim_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n}\tag{1}$$ exists and hence defines a function of $x$ say $F(x)$. This function is called the exponential function and traditionally denoted by symbol $\exp(x)$ or $e^{x}$. But we stick to $F(x)$ for the time being. Next step is to show that $F(x + y) = F(x)F(y)$ for all real $x, y$. And the last part is to show that $$\lim_{x \to 0}\frac{F(x) - 1}{x} = 1\tag{2}$$ This is not difficult but at the same time not obvious.

The proof for $(2)$ goes roughly as follows. First we show that $F(x) \geq 1 + x$ for all $x \in (-1, \infty)$ and from this we can obtain the inequality $$1 \leq \frac{F(x) - 1}{x} \leq \frac{1}{1 - x}$$ for $0 < x < 1$ and then by Squeeze theorem we get the limit $(2)$ for the case when $x \to 0^{+}$. The result for $x \to 0^{-}$ follows by noting that $F(x)F(-x) = 1$ for all $x$.

I have provided an outline for my approach above. for more details see this answer.