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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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    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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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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    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.2011-06-09
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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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    It quite depends on your experiences. I think the curriculum system in many universities can be a good reference for answering your question.2011-06-09
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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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    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.2011-06-08