I am almost certain the book is wrong on this "proof" of a limit. 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 $\left\lvert \frac{1}{x-1} \right\rvert > M$ for $ 0 < \vert x -1 \vert < \delta$. 
Since $\left\lvert \frac{1}{x-1} \right\rvert > M $ is equivalent to $\left\lvert x -1 \right\rvert < \frac{1}{M}$, take $\delta = \frac{1}{M}$ Then for all $x$ satisfying $0 < \vert x-1 \vert < \delta = \frac{1}{M}$, we have $\left\lvert \frac{1}{x-1}\right\rvert > M$ 
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 sufficient 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 quantity. 
Definition 2  Suppose that we have a function $f$ fancy looking one sorry cannot find the LaTeX command for that : $\mathring{U}\left(x_0\right) \to \mathbb{R}$. 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 $x \to x_0$, denoted by
$$\lim_{x\to x_0} f(x) = \infty\,\mbox{ or } f(x) \to \infty \mbox{ as } x\to x_0 $$
If we use $f(x) > M$ (or $f(x) < -M$) instead of $\vert f(x) \vert >M$ in the above definition then $f(x)$ is called a positive (or negative) infinity as $x \to x_0$, denoted by 
$$\lim_{x\to x_0} f(x) = +\infty \left(\mbox{ or } \lim_{x\to x_0} f(x) = -\infty\right)$$

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? 
 A: 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.
A: 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. 
A: 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.]
A: Given their Definitions and the proof, there is NO error in the proof. Any line of reasoning that something is off, would be directed at the definition given. The definition as an axiom, the proof is a logical consequence.
A better question would be about their definitions. 
