Proving this limit as $x \rightarrow ∞$?

Question

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.

Answer 1

Consider $x>0$ and start from here to find $\delta$

$$\left|\frac{x}{x+1}-1\right|<ε\iff \left|\frac{-1}{x+1}\right|<ε\iff 0<\frac{1}{x+1}<ε$$

$$\iff x+1>\frac{1}{ε} \iff x>\frac{1}{ε}-1=\delta $$

Answer 2

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