convergence and limits How can I rewrite $\frac{1}{1+nx}$ and prove it's absolute convergence as $n \rightarrow \infty$? Given $\epsilon > 0$, should I define $f_n(x)$ and $f(x)$? 
Any help is hugely appreciated. Thank you

First decide on what the answer will be. Let's not forget about $x=0$, the limit then is $1$. In all other cases, it is intuitively clear that the limit is $0$. If $x$ is negative, we may need to not start at $n=1$, in order to avoid division by $0$.
Now for the $\epsilon$ stuff. We look separately at the cases $x\gt 0$ and $x\lt 0$, and leave $x=0$ to you. 
Fix $x\gt 0$.  In order to make $\frac{1}{1+nx}\lt \epsilon$, it is enough to make sure that $1+nx\gt \frac{1}{\epsilon}$. This will certainly be true if $nx\gt \frac{1}{\epsilon}$, that is, if $n \gt \frac{1}{x\epsilon}$. So if for example we pick $N =\left\lfloor\frac{1}{x\epsilon}\right\rfloor$, then for every $n\gt N$ we will have $\frac{1}{1+nx}\lt \epsilon$.
Now we deal with the more challenging case where $x\lt 0$. Because negative numbers can cause confusion, replace $x$ by $-y$.  So $y=|x|$. Then our general term is $\frac{1}{1-ny}$. If $ny\gt 1$, the absolute value of this is $\frac{1}{ny-1}$. We will want to make sure that $ny \gt 1$, so we will (among other things) ask that $n \gt\frac{1}{y}$.
We will also want $\frac{1}{ny-1}\lt \epsilon$. Sinc everything is positive, we can rewrite this as $ny \gt 1+\frac{1}{\epsilon}$, and then as $n\gt \frac{1}{y}+\frac{1}{y\epsilon}$. 
Now let $N$ be the greater of $\left\lfloor \frac{1}{y}\right\rfloor$ and 
$\left\lfloor \frac{1}{y}+\frac{1}{y\epsilon}\right\rfloor$. If $n\gt N$, then
$\left|\frac{1}{1+nx}\right|\lt \epsilon$. 
Remark: We have given a very formal argument, supplying some intuition, but perhaps not enough. If some parts are a little too dense, they can be undensified. 
