Prove that $-\frac {x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$ where $f(x)= \frac{1-e^{-x}}{x}$ Let $$\begin{cases}
f(x)= \frac{1-e^{-x}}{x},  & \text{if $x > 0$} \\
f(0)=1
\end{cases}$$
I have to prove that $\forall x \in \Bbb{R_+}-\frac {x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$ so i can show that $f'_d(0)=-\frac{1}{2}$ 
I tried using the mean value inequality but it didn't get me anywhere.
 A: The given inequality is equivalent to: for $x>0$,
$$x\left(1-\frac {x}{2}\right)\leq 1-e^{-x}\leq x\left(1+\frac{x^2}{6}-\frac{x}{2}\right),$$
that is
$$1-x\left(1+\frac{x^2}{6}-\frac{x}{2}\right)\leq e^{-x}\leq 1-x\left(1-\frac {x}{2}\right)$$
or
$$1-x+\frac {x^2}{2}-\frac{x^3}{6}\leq e^{-x}\leq 1-x+\frac {x^2}{2}.$$
Now, you may apply the Lagrange's Remainder of the Taylor's expansion of $e^{-x}$ centered at $x_0=0$:  $\exists t,s\in (0,x)$ such that
$$e^{-x}=1-x+\frac {x^2}{2}-\frac{t^3}{6}=
1-x+\frac {x^2}{2}-\frac{x^3}{6}+\frac{s^4}{24}.$$
Can you take it from here? 
A: We know that Taylor series for $e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}...$ now $f(x)\approx \frac{1-(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24})}{x}=\frac{x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}}{x}=1-\frac{x}{2}+\frac{x^2}{6}-\frac{x^3}{24}$ thus $-\frac{x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$
A: For $x\ge0$, the given inequation is equivalent to 
$$1-x+\frac {x^2}{2}-\frac{x^3}{6}\leq e^{-x}\leq1-x+\frac{x^2}{2}.$$
As equalities hold at $x=0$, you can differentiate and this is equivalent to
$$-1+x-\frac{x^2}{2}\leq -e^{-x}\leq-1+x.$$
As equalities hold at $x=0$, you can differentiate and this is equivalent to
$$1-x\leq e^{-x}\leq1.$$
As equalities hold at $x=0$, you can differentiate and this is equivalent to
$$-1\leq-e^{-x}\leq0.$$
More generally, $e^{-x}$ is bracketed between successive Taylor approximations.

