Let $x\in[0,\infty)$ and a sequence $(a_n)_{n\in \mathbb{N}}$ given by $$\frac{1-xn}{1+xn}$$ Show that the sequence converges and determine the limit as a function of $x$.
Attempt
I do know the definition for convergence but not sure on how to (or even if to) use it to show convergence here. Do I have to reduce my $n$ (I know that doesn't show anything, though)? Is there any test to use that I am not seeing which I can use and then argue for $x$ in the given interval?
If $x=0$, then this is trivial, as $a_n(x)=1$.
Notice that for $x\neq 0$, $$\begin{align*} a_n(x) &=\frac{1-xn}{1+xn}\\ &=\frac{1+xn-xn-xn}{1+xn}\\ &=1-\frac{2xn}{1+xn}\\ &=1-\frac{2x}{(1/n)+x}\end{align*}$$ so we clearly have $$\lim\limits_{n\to\infty}a_n(x)=1-\frac{2x}{x}=-1 $$