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.

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 $-\frac{x}{2}\leq f(x)-1\leq \frac{x^2}{6}-\frac{x}{2}$
