#### Answer

$e^{-x}\approx1-x+\frac{x^2}{2}$ for all x close to zero

#### Work Step by Step

The Taylor series for $e^{-x}$ with $r$ in $(-\infty,\infty)$ is
$1+(-x)+\frac{1}{2!}(-x)^2+\frac{1}{3!}(-x)^3+...+\frac{1}{n!}(-x)^n+...$
$=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...+\frac{1}{n!}(-x)^n+...$
We can use the first few terms of a Taylor Series to get an approximate value for a function.
Hence, we can suggest that $e^{-x}\approx1-x+\frac{x^2}{2}$ for all x close to zero