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Advanced Mathematics by Mingming Chen, Zhengyou Guo Jingxian Yu, Jinqiu Li. Chemical Industry Press pg 28, section 1.4.2 Example 2.

Prove $$\lim_{x \to 1} \frac{1}{x-1} = \infty$$

Proof $\;\forall\, M > 0$, we want to find $\delta > 0$ such that $\vert \frac{1}{x-1} \vert > M$ for $ 0 < \vert x -1 \vert < \delta$.

Since $\vert \frac{1}{x-1} \vert > M $ is equivalence to $\vert x -1 \vert < \frac{1}{M}$, take $\delta = \frac{1}{M}$ Then for all $x$ satisfying $0 < \vert x-1 \vert < \delta = \frac{1}{M}$, we have $\vert \frac{1}{x-1}\vert > M$

Therefore, $\lim_{x \to 1} \frac{1}{x-1} = \infty$


requested by Chris Culter

1.4.2 Infinity Quantity,

Definition 1: If the limit of a function $f\left(x\right)$ as $x \rightarrow x_0$ (or $x \rightarrow \infty$) is 0. Then the function $f\left(x\right)$ is called an infinitesimal quantity with respect to $x\rightarrow x_0$ (or $x\rightarrow \infty$).

Theorem 1 The necessary and sufficiten condition for lim $f\left(x\right) = A$ is $f\left(x\right) = A + \alpha\left(x\right)$, where $\alpha\left(x\right)$ is an infinitesimal.

Definition 2 Suppose that we have a function $f$ fancy looking one sorry cannot find the latex command for that : $\mathring{U}\left(x_0\right) - \mathbb{R}$. If $\,\forall\, M >0$, $\exists \, \delta >0$, such that $\vert f\left(x\right) \vert > M$ for all $x$ satisfying $0 < \vert x-x_0 \vert < \delta$, then $f\left(x\right)$ is called an infinity as $x \rightarrow x_0$, denoted by $$\lim_{x\to x_0} f\left(x\right) = \infty\,\mbox{ or } f\left(x\right) \rightarrow \infty \mbox{ as } x\rightarrow x_0 $$

If we use $f\left(x\right) > M$( or $f\left(x\right) < -M$) instead of $\vert f\left(x\right) \vert >M$ in the above definition then $f\left(x\right)$ is called a positive (or negative) infinity as $x \rightarrow x_0$, denoted by $$\lim_{x\to x_0} f\left(x\right) = +\infty \left(\mbox{ or } \lim_{x\to x_0} f\left(x\right) = -\infty\right)$$


I am confused because $$\lim_{x \to 1^-} \frac{1}{x-1} \neq \lim_{x \to 1^+} \frac{1}{x-1}$$ So then the limit is DNE

Did I miss something?

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typo alert! Have you checked the eratta (if it exists)? –  Gamma Function Sep 4 '13 at 2:58
    
@JacobMayle I tried searching for that, but none are to be found, most book in my country that are written in English are filled with errors, so I ask here when I am unsure. Haven't dealt much with the error, distance def. of limits. –  yiyi Sep 4 '13 at 3:02
    
@yiyi Somewhere in the first 28 pages of this book, there should be a definition of what it means for a limit to equal $\infty$. Could you please find the definition and quote it here? –  Chris Culter Sep 4 '13 at 3:05
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Maybe they have defined with they mean by unsigned infinity? –  Pedro Tamaroff Sep 4 '13 at 3:17
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@yiyi: the definitions reveal that the proof is correct - see the edit to my answer. –  Anthony Carapetis Sep 4 '13 at 6:44

3 Answers 3

up vote 5 down vote accepted

This depends on how you're extending $\mathbb{R}$ to "infinity" - if you're using the extended real line (with the two infinities $\pm \infty$ with respective neighbourhoods $\pm x >M$) then you are correct. However, if you're using the one-point/projective compactification of the real line, then there is only one point at infinity (with neighbourhoods $|x| > M$) and the given proof is correct.

From the definitions you added in your edit we can see that this author uses $\infty$ to mean the single infinity in the one-point compactification, and separately $\pm \infty$ to mean the two infinities in the extended real line. Thus the proof is indeed correct.

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2  
This one point compactification business isn't really what's done in elementary calculus. According to how we usually work at that level, limits are assumed to be in the real numbers, extended by two points, which are denoted $\infty$ and $-\infty$ –  G Tony Jacobs Sep 4 '13 at 3:04
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@GTonyJacobs: Absolutely, and I think it's just as likely that it's a mistake in the text as it is that the text actually uses the one-point compactification. It's just that the proof and notation used within is completely consistent with this. We'd have to know the book to be sure. –  Anthony Carapetis Sep 4 '13 at 3:07
    
It's also possible (see other answer) that only the two-point compactification is used, and unsigned $\infty$ is not used on its own as a value, but appears only in equations $\dots =\infty$ as an abbreviation of the words "... is infinite". In this interpretation there is $+\infty = \infty$ and $-\infty = \infty$ but not $-\infty = +\infty$. –  zyx Sep 4 '13 at 7:18
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@zyx: right. The one-point compactification is the way to make this "limit" an actual limit in a topological space, so that the equality is actually equality. Of course the author's notation doesn't distinguish which space is being used so interpreting $\lim$ as a well-defined operation already has some issues. –  Anthony Carapetis Sep 4 '13 at 7:43
    
It doesn't look to me like a context where $-\infty$ and $+\infty$ are equal is philosophically compatible with simultaneous use of epsilons, inequalities, and estimates. The value and the nature of the unsigned "limit" seem to be qualitative, consistent with equivalent uses of language about a function having an asymptote or singularity at a point. It is possible to have limit as a well-defined operation by considering the set of limiting values, which is a good thing to do for other reasons, but it may not be necessary if the language of (unsigned) limits is used only as a metaphor. –  zyx Sep 4 '13 at 8:22

You are right, the limit is usually considered not to be $\infty$. The limit as $x$ approaches $0$ from the left is $-\infty$. That is the point of view taken in all the calculus books I have used as texts. Identifying $+\infty$ and $-\infty$ is highly unhelpful from the point of view of graphing.

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What about the comment in Anthony carapetis answer? The extending the difference between the extended real line vs. one-point/projective compactification of the real line? –  yiyi Sep 4 '13 at 3:04
    
@yiyi: I have the same experience as André that almost all calculus texts use the extended real line. You should read your text in detail to see what definition they are using. –  Anthony Carapetis Sep 4 '13 at 3:05
    
@yiyi: The conventions are different in different areas of mathematics. When we "extend" the Euclidean plane to the projective plane, we add a whole line of "points at infinity." In the theory of complex variables, extending by adding a single "point at infinity" is useful. In calculus/analysis in one real variable, the useful thing is the familiar $+\infty$, $-\infty$ duo. –  André Nicolas Sep 4 '13 at 7:03
    
There is an interesting situation with this book, in that they use signed expressions $\lim = \pm \infty$ in the ordinary sense, but the unsigned $\infty$ is a metaphor and the definition of $\lim = \infty$ appears to be there only as formal support for the imagery of something being "infinite" (as in "singular", "blowing up", having an asymptote, or other qualitative expressions). If that's true then one-point compactification would be taking the metaphor literally when they mean it poetically! –  zyx Sep 4 '13 at 19:57

The inequalities in the proof refer to $|x-1|$ and not $(x-1)$. Therefore, the argument can be read as a correct proof of the limit of $\frac{1}{|x-1|}$, or as an error of missing absolute value signs "$| \cdots |$" in the statement of the result.

One-point compactification is used less in real analysis, and more in complex analysis or with rational functions. This function is rational, but there is no need for inequalities to compute limits of rational functions in reduced form. Substitution of $1$ for $x$ can be done directly. If the book means the limit for complex $x \to 1$ then the statement and proof can be considered correct.

[The Update has clarified things. In the book's definition, $\lim f = \infty$ is the same as $\lim |f|=+\infty$. That resolves the contradiction, and shows why the inequalities use $|x−1|$. It is not uncommon to say things like $\sin(x)/x^2$ "is infinite" at $x=0$, although the sign is different on the two sides. In accomodating this kind of language, the book is using $\pm \infty$ as numerical values (in an extended real number system) and "$=\infty$" as a property that a value can have. This allows for the equation $-\infty = \infty$, while avoiding the interpretation that one- and two-point compactifications are both being used at the same time on the same page of the book.]

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I have added the writers names also the isbn: 978-7-122-09459-9 –  yiyi Sep 4 '13 at 6:09
    
Thanks. I edited the answer. –  zyx Sep 4 '13 at 7:01

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