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.

It sounds awkard to use these inequalities to prove the limit you want.
The limit $\lim_{x\to 0} (e^x-1)/x=1$ follows by L'Hospital's rule as you said, or by using the fact that $(e^x-1)/x=1+\sum_{n\ge 1} x^n/(n+1)!$ and that the power series $\sum_{n\ge 1} x^n/(n+1)!$ is absolutely convergent and hence the sum converges to $0$ as $x\to 0$. If you are not convinced of that at sight you may use  $|\sum_{n\ge 1} x^n/(n+1)!|\le |x|\sum_{n\ge 0} |x|^n/(n+2)!\le |x|e^{|x|}$.
Now to prove the inequality $(e^x-1)/x\le 1+|x|(e-2)$ here are some hints. First note that the case $x\in [-1,0[$ follows from the case $x\in ]0,1]$, since for $x<0$ we have $(e^x-1)/x=(e^{-|x|}-1)/(-|x|)=(1-e^{-|x|})/|x|=(e^{|x|}-1)/(|x|e^{|x|})\le (e^{|x|}-1)/|x|$, since $e^u\ge 1$ if $u\ge 0$. Now let's concentrate on the case $x\in I=]0,1]$. In that case we may write
$(e^x-1)/x=1+x\sum_{n\ge 0} x^n/(n+2)!\le 1+x\sum_{n\ge 0}1/(n+2)!=1+x(e-2)$.
I didn't try, but I guess similar arguments may handle the other inequality. I hope this helps.
