Proving this limit as $x \rightarrow ∞$? I'm having trouble proving this using epsilon-delta:
$$\lim_{x\to\infty} \left|\frac{x}{x+1}\right|=1$$
I translated this into:
$$\forall \epsilon>0,∃\delta\in\mathbb R,x>\delta\implies \left|\frac{x}{x+1}-1\right|<\epsilon$$
I don't really know where to go from here. Any help would be appreciated.

Try observing that
$$
\left|\frac{x}{x+1}-1\right|=\left|\frac{x-x-1}{x+1}\right|
=\left|\frac{1}{x+1}\right|
$$
