<s>[INST] Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$. Could someone please explain a step in the following proof?

Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.

The above limit exists if for every $\varepsilon > 0$, there exists a real number $\delta > 0$ such that if $0<|x-1|<\delta$, then $\left |\frac{1}{x+1}-\frac{1}{2}  \right |=\left | \frac{2-x-1}{2(x+1)} \right |=\left | \frac{1-x}{2(x+1)} \right |=\frac{|1-x|}{2|x+1|}=\frac{|x-1|}{2|x+1|}<\varepsilon$.
I understand that $|x-1|$ can be made small, but for $|x+1|$, why does the solution say the condition "if $|x-1| < 1$, $x+1\in (1,3)$"? Why and how is the condition $|x-1| <1$ made? 
 [/INST] So far we have $$\left |\frac{1}{x+1}-\frac{1}{2}  \right |=\left | \frac{2-x-1}{2(x+1)} \right |=\left | \frac{1-x}{2(x+1)} \right |=\frac{|1-x|}{2|x+1|}=\frac{|x-1|}{2|x+1|}<\varepsilon$$ In order to achieve the last inequality, we have to control the denominator as well as the numerator. 
By assuming $|x-1|<1$ we can control $|x+1|$, therefore the fraction, $ \frac{|x-1|}{2|x+1|}$ could be made less than $\epsilon$ by choosing a small enough $ \delta$ .
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