Taylor expansion of $f(x) = \frac{x}{x+3}\frac{1}{x-2}$ near $x=2$. I am trying to Taylor expand the function $$f(x) = \frac{x}{x+3}\frac{1}{x-2}$$ around the point $x_0 = 2$. Clearly, the last factor explodes around this point, so I will try and expand that term. However, terms on the form $\frac{1}{x-2}$ are a geometric series. Is that simply what I have to use?

I think here we may rather talk about a Laurent series expansion around $x=2$.
We have, as $x \to 2$,
$$
\begin{align}
f(x) &= \frac{1}{x-2}\frac{x}{x+3}\\\\
&= \frac{1}{5(x-2)}\frac{2+(x-2)}{\left(1+\dfrac{(x-2)}5\right)}\\\\
&= \frac{2+(x-2)}{5(x-2)}\sum_{n\geq0}\dfrac{(-1)^n}{5^n}\left(x-2\right)^n,\quad 0<|x-2|<5,\\\\
&= \left(\frac15+\frac{2}{5(x-2)}\right)\sum_{n\geq0}\dfrac{(-1)^n}{5^n}\left(x-2\right)^n\\\\
&=\frac25\frac{1}{(x-2)}+\frac{3}{25}-\frac{3(x-2)}{125}+\frac{1}{625} (x-2)^2-\frac{(x-2)^3}{3125}+\cdots\\\\
&=\frac25\frac{1}{(x-2)}+3\sum_{n\geq0}\dfrac{(-1)^n}{5^{n+2}}\left(x-2\right)^n.
\end{align}
$$
