# How do you find the Taylor remainder term R_3(x;0) for f(x)=1/(2+x)?

This exact value of ${R}_{n} \left(x\right)$ is

for some value of $c$ between $a$ and $x$

However we can determine an upper bound for ${R}_{3} \left(x\right)$

which is

Hence for $a = 0$ and $n = 3$ we get

where

hence

${f}^{4} \left(0\right) = \frac{24}{2} ^ 5 = \frac{3}{4}$

So finally we have that