Error of Taylor Polynomial a) Construct the Taylor polynomials of degree 4 at $x_0 = 0$ for the following functions:
$$f(x) = \frac{1}{2+x}$$$$ f(x)=\sin\left(\frac{x}{3}\right)$$
b) Find a bound for the error terms for $x\in [-1,1].$
I have the solution to the first part as 
$$\frac{1}{2+x} \approx \frac{1}{2}-\frac{x}{4} +\frac{x^2}{8} -\frac{x^4}{32} $$
and
$$\sin\left(\frac{x}{3}\right) \approx \frac{x}{3}-\frac{x^3}{162}$$ but I am completely stumped at finding the error bounds in part b.

Recall the remainder term is bounded by $\displaystyle{Mr^{n+1}\over (n+1)!}$ where $M=\displaystyle\sup_{a\le x\le b}|f^{(n+1)}(x)|$ and $r={b-a\over 2}$ is the radius of the interval in question. In your case you know the derivatives well enough to see that
$$|f^{(n+1)}(x)|=\begin{cases}\displaystyle{n!\over (2+x)^{n+1}} && f(x) = (x+2)^{-1} \\ 3^{-n-1}|g(x)| && f(x) = \sin(x/3)\end{cases}$$
Where here $g(x)$ is something bounded by $1$, but would be a sine or cosine depending on $n$ and I'm too lazy to write that out. So you get bounds
The maximum of the first is just $n!$ since the smallest the denominator can be is $2+-1=1$. For the second we just bound by $1$ for simplicity. This gives rise to error bounds

$$B_1={1\over n+1}$$
  $$B_2={1\over 3^{n+1}(n+1)!}$$

