<s>[INST] Prove using Taylor's Theorem: $1-\cos(\frac{x}{n})=\frac{x^2}{n^2}-\cos\theta(x_n,n)\frac{x^3}{6n^3}$ let $F_n(x)=n^2(1-\cos(\frac{x}{n}))$. I am asked to prove that this converges pointwise to limit function $F$ on a set $S$. In this case using geogebra, I see that it converges to $F=x^2$ where $x\in \mathbb{R}$.
From the solutions manual, I found they got this using Taylor's theorem,
$1-\cos(\frac{x}{n})=\frac{x^2}{n^2}-\cos\theta(x_n,n)\frac{x^3}{6n^3}$
where $\theta(x_n,n)$ is between $0$ and $\frac{x}{n}$
Furthermore, they state that $|F_n(x)-x^2|\leq \frac{x^4}{24n^2}$.
How can I show, 
$1-\cos(\frac{x}{n})=\frac{x^2}{n^2}-\cos\theta(x_n,n)\frac{x^3}{6n^3}$
and $|F_n(x)-x^2|\leq \frac{x^4}{24n^2}$.
What I have so far:
$1-\cos(\frac{x}{n})=1-\cos(\frac{a}{n})+\sin(\frac{a}{n})*(\frac{1}{n})*(\frac{x}{n}-\frac{a}{n})^2+\cos(\frac{a}{n})*(\frac{1}{n})*(\frac{x}{n}-\frac{a}{n})^3$....  
How would I proceed.
 [/INST] Hint. Assume $h>0$. Integrating by parts three times, one has
$$
\begin{align}
\cos h-1&=\int_0^h-\sin t\:dt
\\&=\left[(h-t) \frac{}{}\sin t\right]_0^h-\int_0^h (h-t)\cos t\:dt
\\&=\color{red}{0}+\left[\frac{(t-h)^2}{2}\cdot\cos t\right]_0^h+\int_0^h \frac{(t-h)^2}{2}\cdot\sin t\:dt
\\&=-\frac{h^2}{2}+\frac12\int_0^h (t-h)^2\cdot\sin t\:dt
\\&=-\frac{h^2}{2}+\left[\frac{(t-h)^3}{6}\cdot\sin t\right]_0^h-\int_0^h \frac{(t-h)^3}{6}\cdot\cos t\:dt
\\&=-\frac{h^2}{2}+\color{red}{0}-\int_0^h \frac{(t-h)^3}{6}\cdot\cos t\:dt
\end{align}
$$ or

$$
\cos h=1-\frac{h^2}{2}+\int_0^h \frac{(h-t)^3}{6}\cdot\cos t\:dt
$$

then one may use the mean value theorem for definite integrals to conclude.
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