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? 

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.
