Is there a proof for $\lim_{x \to a} \frac{1}{x-a} = \infty$? I am an adult software developer who is trying to do a math reboot. I am working through the exercises in the following book.
Ayres, Frank , Jr. and Elliott Mendelson. 2013. Schaum's Outlines Calculus Sixth Edition (1,105 fully solved problems, 30 problem-solving videos online). New York: McGraw Hill. ISBN 978-0-07-179553-1.
So far as I can tell, the following question either has a misprint or the book does not cover the material.  It is entirely possible that I failed to grasp a key important sentence.
Chapter 7 Limits, problem 24.
Use the precise definition to prove:
$$
\text{a)  }\lim_{x \to 0} \frac{1}{x} = \infty \\
\text{b)  }\lim_{x \to 1} \frac{x}{x-1} = \infty \\
$$
My understanding.
It is possible to prove $\lim_{x \to 0^+} \frac{1}{x} = +\infty$ or $\lim_{x \to 0^-} \frac{1}{x} = -\infty$, but not $\lim_{x \to 0} \frac{1}{x} = \infty$ because $\frac{1}{x}$ is a hyperbola with no limit at 0.  A similar argument can be made for $\frac{x}{x-1}$ at 1.
Is there a proof for $\lim_{x \to a} \frac{1}{x-a} = \infty$?

A function is not defined properly unless you specify its domain. When we say the function $\frac{1}{x}$ we usually mean the function $f: \mathbb{R}\setminus \lbrace 0 \rbrace \rightarrow \mathbb{R} $ defined by $f(x)=\frac{1}{x}$ for all $x \in \mathbb{R} \setminus \lbrace 0 \rbrace $. Here $\lim_{x \to a} f(x) $ does not eexist.
But when you take $g=f_{|[0, \infty)}$ then we can say that $\lim_{x \to a} g(x) = \infty$.
So try to find out the domain of the mapping from the context.
