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? 

Limits basically come as equivalent to the shadow (or standard part) function "sh" of non-standard analysis.  So, consider 1+e where e represents an infinitesimal.  Then 
sh((1+e)(1+e)+1)=sh(1+2e+ee+1)=sh(2+2e+ee)=sh(2)+sh(2e)+sh(ee)=2+0+0=2.
