- Context
I am reading the textbook Calculus With Applications, by Peter D. Lax and having a problem understanding the proof of $\frac{\mathrm{d}e^x}{\mathrm{d}x}\vert_{x=0}=1$, the result of which is then used to prove the derivative of $e^x$.
The textbook first uses the inequalities $$ \left(1+\frac{1}{n}\right)^n<e<\left(1+\frac{1}{n}\right)^{n+1}<\left(1+\frac{1}{n-1}\right)^n \tag{1}\label{eq1} $$ for all integers $n>1$. Then it takes $h=\frac{1}{n}$ as a special sequence of $h$ tending to zero. After a few derivations of the inequality, it get $$ 1\le\frac{e^h-1}{h}\le\frac{n}{n-1} $$ As n tends to infinity, the right-hand term tends to 1, so by the squeeze theorem, the center term tends to 1 also, which means the derivative of $e^x$ at $x=0$ is equal to 1.
However, as it says, this does not quite finish the proof, since it has taken $h$ to be of the special form $h=\frac{1}{n}$. So it leaves a problem to fill this gap:
- Complete problem description
Recall from Eq.$\eqref{eq1}$ that e is between the increasing sequence $e_n$ and the decreasing sequence $f_n$. Explain the following items.
(a) If $h>0$ is not of the form $\frac{1}{n}$, then there is an integer $n$ for which $\frac{1}{n}<h<\frac{1}{n-1}$.
(b) Using Eq.$\eqref{eq1}$, one has $\left(1+\frac{1}{n}\right)^{nh}<e^h<\left(1+\frac{1}{n-2}\right)^{\left(n-1\right)h}$.
(c) $\left(n-1\right)h<1<nh$
(d) $1+\frac{1}{n}<e^h<1+\frac{1}{n-2}$
(e) $n-1<\frac{1}{h}<n$, and $\frac{n-1}{n}<\frac{e^h-1}{h}<\frac{n}{n-2}$.
Conclude from this that $\frac{e^h-1}{h}$ tends to $1$ as $h$ tends to zero.
- My progress
I have two main questions, the first one is how to prove (a) and the second one is how the final conclusion is obtained from (e). And I have proved the intermediate (b), (c), (d) and (e) process.