Finding the radius of convergence for $\frac{1}{(3 - x)(1 + x)^{2}}$ at $x_{0} = 0$ and $x_{0} = 1$ I am having trouble with the problem 

Find the radius of convergence for $f(x) = \frac{1}{(3 - x)(1 +
 x)^{2}}$ at $x_{0} = 0$ and $x_{1} = 1$.

Usually, when I'm trying to find radius of convergence, I'm given some sort of series (like, a summation). This isn't the case here, and I'm confused as to what I'm supposed to do. This is a problem in the exercise. It is in the section for power series. Also, I have this formula for the radius of convergence $R$:
$$R = \frac{1}{\limsup_{n\to\infty} |a_{n}|^{1/n}},$$
but I'm not sure how I'm supposed to use that here either. 
Any help is appreciated.
EDIT: Maybe it has something to do with the standard Taylor series form? Like, 
$$f(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + f''(x_{0})(x - x_{0})^{2}/2! \ldots $$
Also, in case it matters, by Partial Fractions, you can write
$$\frac{1}{(3-x)(1 + x)^{2}} = \frac{1}{16(x + 1)} + \frac{1}{4(x+1)^{2}} - \frac{1}{16(x - 3)}. $$
The first few terms of the Taylor series expansion about $x_{0} = 0$ is given by
$$f(x) = \frac{1}{3} - \frac{5}{9}x + \frac{22}{27}x^{2} - \frac{86}{81}x^{3} + \frac{319}{243}x^{4} + \cdots $$
(I don't have a closed term for the sum, but these are the first terms explicitly computed. Looks like the denominator is just powers of $3$. Not sure about the numerator though...)
 A: If you represent $\frac {1}{(3-x)(1+x)^2}$ as a Taylor series, centered at the points identified above that Taylor series will have a finite radius.  There is enough information presented in the question to find these radii without deriving the entire series.
However, if you represent the function as a Laurent series, you have more choices in how you represent the series and depending on those choices the series will converge in different annuli. 
The trick comes down to finding the discontinuities of the function.  Since the Taylor series is continuous, it is going to diverge as these discontinuities approach.
A: First, write the given series as a power one around some given point, then evaluate the radius of convergence.
For example, about $\;x=1\;$ :
$$\frac1{(3-x)(x+1)^2}=-\frac1{16}\frac1{x-3}+\frac1{16}\frac1{x+1}+\frac14\frac1{(x+1)^2}=$$
$$=\frac1{32}\frac1{1-\frac{x-1}2}+\frac1{32}\frac1{1+\frac{x-1}2}+\frac1{16}\frac1{\left(1+\frac{x-1}2\right)^2}=$$$${}$$
$$=\frac1{32}\left[\sum_{n=0}^\infty(-1)^n\frac{(x-1)^n}{2^n}+\sum_{n=0}^\infty\frac{(x-1)^n}{2^n}+2\left(\sum_{n=1}^\infty(-1)^n\frac{(x-1)^n}{2^n}\right)^2\right]$$
Now, if you had to write down a power series then that'd be really tedious, though not quite difficult, but you're only required to calculate the convergence radius, and that seems pretty easy as all the three series' convergence radius above is $\;1\;$ around the point we want, so it must be
$$\left|\frac{x-1}2\right|<1\iff |x-1|<2$$
Note that in fact all we needed above is what follows after the simple fractions decomposition, as we already there have the basic $\;\frac{x-1}2\;$ thing we shall need. To write down the infinite series is just for fun...
A: Here are my thoughts. This might not be the original purpose of the exercise, but this is my guess.
According to your partial fraction decomposition, and using the well-know power series $(1-x)^{-1} = \sum_0^\infty x^n$ for $x \in (-1,1)$, at $x=0$, we have
\begin{align*}
f(x)&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac {\mathrm d}{\mathrm dx} \left( \frac 1{4(x+1)} \right) + \frac 1{48} \cdot \frac 1{1 - x/3} \\
&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac 14 \cdot\frac {\mathrm d }{\mathrm dx}\left( \sum_0^\infty (-1)^n x^n \right) + \frac 1{48} \cdot \sum_0^\infty \frac {x^n}{3^n}\\
&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \frac 14  \sum_0^\infty (n+1)(-1)^{n+1} x^{n}  + \frac 1{48} \cdot \sum_0^\infty \frac {x^n}{3^n}\\
&= \sum_0^\infty \left(\frac {(-1)^n}{16} + \frac 14 (n+1)(-1)^{n+1} +\frac 1{16 \cdot 3^{n+1}} \right) x^n,
\end{align*}
and all of these computations are valid iff $|x|<1$ and $|x/3|<1$. Conclusively the radius  is $1$. You might try to compute again by the formula. 
At $x=1$, just let $y = x-1$, then $x = y+1$ and 
$$
f(x) =g(y) = \frac 1{(2-y)(2+y)^2},
$$ 
you could do the similar thing like above to give the power series at $y=0$. You could see that the operations are valid iff $|y/2|<1$, hence $|y|<2$ and the radius is $2$. 
