Answer
$${\text{ }}\frac{1}{{1 + x}} \approx 1 - x$$
Work Step by Step
$$\eqalign{
& \frac{1}{{1 + x}} \approx 1 - x,{\text{ }}{x_0} = 0 \cr
& {\text{Let }}f\left( x \right) = \frac{1}{{1 + x}} \cr
& {\text{ }}f\left( {{x_0}} \right) = f\left( 0 \right) = \frac{1}{{1 + 0}} = 1 \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{1 + x}}} \right] \cr
& f'\left( x \right) = - \frac{1}{{{{\left( {1 + x} \right)}^2}}} \cr
& f'\left( {{x_0}} \right) = - \frac{1}{{{{\left( {1 + 0} \right)}^2}}} = - 1 \cr
& {\text{We can approximate values of }}f\left( x \right){\text{ by}} \cr
& {\text{ }}f\left( x \right) \approx f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \cr
& {\text{then}} \cr
& {\text{ }}\frac{1}{{1 + x}} \approx 1 + \left( { - 1} \right)\left( {x - 0} \right) \cr
& {\text{ }}\frac{1}{{1 + x}} \approx 1 - x \cr} $$