Proving that $e^x = 1+O(x)$ as $x \to 0$.

Question

I'm trying to recall how to rigorously show that $e^x = 1 + O(x)$ as $x \to 0$.

I believe one way is to use Taylor's theorem applied to the power series expansion of $e^x$. We have

$$e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

The Lagrange form of Taylor's theorem tells us that for all $x$ there exists a $c$ between $0$ and $x$ such that $e^x = 1 + e^cx$.

So if $x \in [-1,1]$, then $e^x = 1 + e^cx \le 1 + 3x$, since $e^c \le 3$ for $c \in [-1,1]$.

This shows that $|e^{x}-1| \le 3|x|$ for all $|x|\le 1$. I believe this completes the proof.

This seems a bit cluttered to me. My questions are:

1. Is it correct?
2. Can it be cleaned up?
3. Is there a way to show it without using Taylor's theorem? Maybe by just looking at the power series expansion?

Answer 1

Hint for an alternative way of showing this: write $f(x) = e^x$. Then \begin{align*} \lim_{h \to 0} \frac{f(h) - 1}{h} = \lim_{h \to 0} \frac{f(h) - f(0)} { h} = f'(0) = f(0) = 1 \end{align*}

Answer 2

If you know that $$e^{x}=\lim_{n\to\infty} \left(1+\frac xn\right)^n$$

Then we have for $x\in [0,1)$ that $e^x\geq 1+x$ and $e^{-x}\geq 1-x$ by Bernoulli's inequality.

But then $e^x<\frac{1}{1-x} = 1+\frac{x}{1-x}$. If $x\in[0,1/2)$ then $1+x<e^x<1+2x$.

When $x\in[0,1/2)$ we already have $1-x<e^{-x}\leq 1$.