Show that $\left|\ln(1+x)-x+\frac{x^2}{2}\right|\le \frac{8|x|^3}{3}$ for $|x| ≤ 1/2$ I first begin by finding the 3-rd derivative of the function $f(x)=\ln(1+x)$ that is $2(1+x)^{-3}$ using Lagrange's remainder form the left side is equal to $\displaystyle \frac{2}{(1+tx)^3}\cdot  \frac{x^3}{3!}$ with $t$ is between 0 and 1, simplified is $\displaystyle\frac{x^3}{3(1+tx)^3}$.
In order for the expression to be as large it can be the numerator should be as large as possible and the denominator as small as possible. I understand why there is a 3 in the denominator in the right expression above but where does the 8 come from?
 A: The $8$ comes from evaluating the "worst" possible value of $1/(1+tx)^3$ with $|x|\le1/2$ and $t$ between $0$ and $1$, as Robert Z's answer explains. But you can prove a better bound, with the $8$ replaced by a $2$, if you take a different approach, using the integral definition of the natural logarithm:
$$\ln(1+x)-x+{x^2\over2}=\int_0^x\left({1\over1+t}-1+t\right)dt=\int_0^x{t^2\over1+t}dt$$
and thus, for $|x|\lt1$, we have
$$\left|\ln(1+x)-x+{x^2\over2}\right|=\left|\int_0^x{t^2\over1+t}dt\right|\le\int_0^{|x|}{t^2\over1-|x|}dt={|x|^3\over3(1-|x|)}$$
Now if $|x|\le1/2$, we have $1-|x|\ge1/2$, and the desired bound follows. 
Note, this approach also uses the "worst" (i.e., smallest) possible value for a denominator. In the derivation here, it does so in two steps, first replacing $1+t$ with $1-|x|$ (which requires $|x|\lt1$), and then replacing $1-|x|$ with $1-(1/2)=1/2$.
A: You are almost done. Just note that for $|x|\leq \frac{1}{2}$ and $t\in(0,1)$,
$$0<\frac{1}{(1+tx)^3}\leq \frac{1}{(1+1\cdot (-\frac{1}{2}))^3}= 8.$$
P.S. If we remove the factor $8$ the inequality does not hold in $x\in (-1/2,0)$: note that $$\frac{\left|\ln(1-1/2)-(-1/2)+\frac{(-1/2)^2}{2}\right|}{|(-1/2)^3|}\approx .545177445>\frac{1}{3}.$$
However, we can replace the constant $8$ with a better bound, for example $2$.
A: This post proves that $ \ln(1+x) \leq x - x^2/2 + x^3/3$ and that for $x > 0$ also $x - x^2/2 \le \ln(1+x)$.
So for $x > 0$  even $\left|\ln(1+x)-x+\frac{x^2}{2}\right|\le \frac{x^3}{3}$ holds. 
For $-\frac12<x<0$ there is extra work to do for the lower limit.
