Trouble with absolute value in limit proof As usual, I'm having trouble, not with the calculus, but the algebra. I'm using Calculus, 9th ed. by Larson and Edwards, which is somewhat known for racing through examples with little explanation of the algebra for those of us who are rusty.
I'm trying to prove $$\lim_{x \to 1}(x^2+1)=2$$ but I get stuck when I get to $|f(x)-L| = |(x^2+1)-2| = |x^2-1| = |x+1||x-1|$. The solution I found says "We have, in the interval (0,2), |x+1|<3, so we choose $\delta=\frac{\epsilon}{3}$." 
I'm not sure where the interval (0,2) comes from. 
Incidentally, can anyone recommend any good supplemental material to go along with this book? 

By the continuity of the function $f(x)=x^2+1$, $$\lim_{x\to 1}(x^2+1)=2$$
For the $\epsilon-\delta$ proof, one needs to show that 
$$\forall \epsilon>0, \exists \delta>0\quad \textrm{s.t.}\quad |x-1|<\delta\Rightarrow |(x^2+1)-2|<\epsilon$$
Now things boil down to finding the $\delta$(depending on $\epsilon$) such that 
$$|(x^2+1)-2|<\epsilon,$$ i.e.,$$|x+1||x-1|<\epsilon.\qquad (*)$$ Note that the choice of $\delta$ is totally decided by you. That's to say, given $\epsilon>0$, you can let any value of $\delta$ such that (*) is satisfied when $|x-1|<\delta$. From $|x-1|<\delta$, we have $1-\delta<|x|<1+\delta$, which implies that 
$$|1+x|\leq 1+|x|\leq 2+\delta.$$
Now we have $$|1+x||1-x|\leq(2+\delta)\delta.$$
If you can make $$(2+\delta)\delta<\epsilon$$ for the given $\epsilon$, things are done. Hence for example, you can chose $\delta$ such that $$0<\delta<\min\{1,\frac{\epsilon}{3}\}.$$  
Since $\delta<1$, $|x-1|<\delta<1$. This is where your $(0,2)$ from.

For the references of the topic, I would strongly recommend Terrence Tao's Real Analysis. 
