Is there a proof for $\lim_{x \to a} \frac{1}{x-a} = \infty$?

Question

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

Answer 1

Also in this case the limit doesn't exist but we can consider the two side limits

$$\lim_{x \to a^+} \frac{1}{x-a} = +\infty$$

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

for the proof a simple way is set $y=x-a\to 0$ and the limits become the simpler

$$\lim_{y \to 0^+} \frac{1}{y} = +\infty$$

$$\lim_{x \to 0^-} \frac{1}{y} = -\infty$$

Answer 2

No since the graph of $\frac1{x-a}$ is just that of $\frac1x$ shifted $a$ units horizontally.

Answer 3

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.