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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?

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    *I'm not sure where the interval (0,2) comes from.* Basically you could choose any interval containing 1. For instance, if you'd choose $(-1,3)$ then you'd get $|x+1|<4$ and you could finish with $\delta=\frac\varepsilon4$. (Or, to be more precise $\delta=\min\{2,\frac\varepsilon4\}$, since $(-1,3)=(1-2,1+2)$.)2011-06-08
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    You have two factors, you want to make their product small. You can make the second factor small. But you need to make the first factor at least bounded. That is the reason for choosing an interval like $(0,2)$ or $(-1,3)$.2011-06-08
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    Eureka! I love that moment when the light turns on. It took a couple read-throughs of everyone's answers and putting the pieces together in my brain so I figured I'd just post this on the original question. Thanks, everyone!2011-06-09

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