# 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?

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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 $(-1,3)=(1-2,1+2)$.) –  Martin Sleziak Jun 8 '11 at 14:51
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 $(-1,3)$. –  GEdgar Jun 8 '11 at 17:57
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! –  Pete Ley Jun 9 '11 at 0:32

Because of the freedom in the choice of $\delta$, you can always assume $\delta < 1$, that implies you can assume $x$ belongs to the interval $(0, 2)$.
Edit: $L$ is the limit of $f(x)$ for $x$ approaching $x_0$, iff for every $\epsilon > 0$ it exists a $\delta_\epsilon > 0$ such that: $$\left\vert f(x) - L\right\vert < \epsilon$$ for each $x$ in the domain of $f$ satisfying $\left\vert x - x_0\right\vert < \delta_\epsilon$. Now if $\delta_\epsilon$ verifies the above condition, the same happens for each $\delta_\epsilon'$ such that $0 < \delta_\epsilon' < \delta_\epsilon$, therefore we can choose $\delta_\epsilon$ arbitrarily small, in particular lesser than 1.

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Sorry, could you explain a little more about why you can always assume that? This book uses that idea in one of the examples but doesn't explain why. –  Pete Ley Jun 8 '11 at 15:21
Ah, silly me. I probably could have reasoned that one out if I had tried. :P I guess I assumed it was more complicated than that. –  Pete Ley Jun 9 '11 at 0:36

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.

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Thanks for a very in depth answer. Just so I don't dive into something over my head, what topics would you recommend having a good understanding of before studying real analysis? –  Pete Ley Jun 9 '11 at 0:38
It quite depends on your experiences. I think the curriculum system in many universities can be a good reference for answering your question. –  Jack Jun 9 '11 at 0:45

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.

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Here's a link on standard part function en.wikipedia.org/wiki/Standard_part However, I am not sure this is a stuff which would be appropriate for beginners. (It would be different if he studied a book which takes a non-standard approach and he learned some rules how to deal with infinitesimals.) –  Martin Sleziak Jun 8 '11 at 15:53
I like nonstandard analysis. I read Robinson's book when I was in college. But I suspect that this answer is of absolutely no help to this poster. –  GEdgar Jun 8 '11 at 18:00