Prove that $1+x \le e^x \le 1+x+x^2$ for every $|x| \le 1$

Question

I get the following inequality formula.

$$1+x \le e^x \le 1+x+x^2 \le \frac1{1-x}\quad\text{if}\quad |x| \le 1$$

I know $\displaystyle e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ but I cannot prove the inequalities for $x\lt0$.

Answer 1

Hint: As we know that $\displaystyle e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

So $e^x \geq 1+x$ as you can delete the part after $1+x$ in the above formula to achieve this part of inequality.

Again we know that $\dfrac{x^2}{2!} \leq x^2$ and so on, hence the right side of the inequality follows.

(for $|x|\leq 1$)