How can we prove this inequality involving e? $$ \frac{1}{1+|x|} \le \frac{e^x - 1}{x} \le1 + |x|(e - 2) $$ where $$x \in [-1,0)\cup(0,1]$$
How can we prove this inequality? My text used it to prove the limit of the middle function when $x→0$ which is $1$ using sandwich theorem. 
I have no idea how to begin. I know I can find the limit using L'Hospital but still I cant figure out where did this inequality come from.

From the series definition it immediately follows that $e^x\geq 1 + x$: by uniqueness of Taylor series, the first derivative at $0$ is $1$, so $1+x$ is tangent - inequality follows by convexity of $e^x$.
1. $\frac{e^x-1}{x}\geq\frac 1{1+|x|}$
If $x>0$, then $e^x\geq 1 + x$ implies
$$\frac{e^x-1}x\geq 1\geq \frac 1{1+|x|}.$$
If $x<0$, then we have 
\begin{align}
\frac{e^x-1}{x}\geq\frac 1{1+|x|}&\iff e^x - 1\leq \frac x{1-x}\\
&\iff e^x \leq 1+\frac x{1-x}\\
&\iff e^x \leq \frac 1{1-x}\\
&\iff 1 - x \leq e^{-x}\\
&\iff 1 + t \leq e^{t}
\end{align}
2. $\frac{e^x-1}{x}\leq 1+|x|(e-2)$
For $x > 0$, we have
\begin{align}\frac{e^x-1}{x}\leq 1+|x|(e-2) &\iff e^x-1\leq x+x^2(e-2)\\
&\iff \frac{e^x-1-x}{x^2}\leq e-2\\
&\iff \frac 1{2!}+\frac x{3!}+\ldots\leq \frac 1{2!} + \frac1{3!}+\ldots\\
\end{align}
If $x<0$, then $$e^x\geq 1 + x \implies e^x-1\geq x\implies \frac{e^x-1}{x}\leq 1 \leq 1+|x|(e-2)$$
