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.
 A: HINT/Partial Solution
Consider the case where $x\lt0$. 
Let's work on the left part inequality $$\frac{1}{1-x}\le\frac{e^x-1}{x}\Rightarrow\\\text{(since $x$ is negative)}\\\frac{x}{1-x}\ge e^x-1\Rightarrow\\e^x(x-1)\le1$$
Let $g(x)=e^x(x-1)$ , $x\in[-1,0)$
We have $$g'(x)=xe^x\lt0, \forall x\in[-1,0)\Rightarrow\\ \text{ g($x$) is increasing in $[-1,0)$ }\Rightarrow\\x\le0\Rightarrow g(x)\le g(0)=1$$
as desired.
For the right part of the inequality, that is $$\frac{e^x-1}{x}\le1-x(e-2)$$
we can work in the same manner, by considering the function $h(x)=e^x-x+x^2(e-2)-1$ (here, you will need to see what happens with $h''(x)$ to find the monotony of $h$). 
The same approach can be applied for the $x\gt0$ case.
A: Power series algebra is your friend here. We have:
$$\frac{1}{1+|x|}=\sum_{i=0}^\infty (-1)^i|x|^i=1-|x|+|x|^2-|x|^3+\cdots$$
$$\frac{e-1}{x}=\sum_{i=0}^\infty \frac{x^i}{(i+1)!}=1+\frac{1}{2}x+\frac{1}{6}x^2+\cdots$$
$$1+|x|(e-2)=1-|x|+x|x|+\frac{1}{2}x^2|x|+\frac{1}{6}x^3|x|+\cdots$$
To see that the first is less than the second, notice that when $1>x>0$ the first is less than $1$ and the second is greater than $1$. When $-1<x<0$, using the Lagrange Error formula tells you that the second is greater than the first. Thus the second is always greater than the first.
See if you can come up with an argument for why the third is greater than the second.
A: 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.
A: 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 $$e^x\geq 1 + x \implies e^x-1\geq x\implies \frac{e^x-1}{x}\leq 1 \leq 1+|x|(e-2)$$
