Use the precise definition to prove $\lim_{x \to 1} \frac{x}{x-1} = \infty$.

Question

This is a concrete follow up to this 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.

The above book does not give an example of solving a question exactly like the following.

Chapter 7 Limits, problem 24.

Use the precise definition to prove:

$$ \text{b) }\lim_{x \to 1} \frac{x}{x-1} = \infty \\ $$

Notes.

Unsigned infinity is defined as such: as $x$ approaches $a$, $|f(x)|$ eventually becomes greater than any preassigned positive number. Hence $\lim_{x \to a} f(x) = \infty$ if and only if $\lim_{x \to a} |f(x)| = +\infty$.

The precise definition of a limit is defined as such: $\lim_{x \to a} f(x) = +\infty$ if and only if, for any positive number $M$, there exists a positive number $\delta$ such that, whenever $0<|x−a|<\delta$, then $f(x)>M$. The definition for $\lim_{x \to a} f(x) = −\infty$ is similar.

My attempt.

Does this work as a proof? I am not convinced this is complete.

Let $0 < M$ be given. Let $0 < \delta = \min\big(1, \frac{1}{|M-1|}\big)$. Suppose $x < 1$ and $0 < |x-1| < \delta$. Then $|x-1| < \frac{1}{|M-1|} < 1$. Hence $|M-1| < \frac{1}{|x-1|}$ or $M < \big|\frac{x}{x-1}\big|$.

Answer 1

Note that we have

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

For the proof we can set $x-1=y\to 0$ and then

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

and then reduce to prove that

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

Answer 2

It looks to me like you are going "backwards". You say "let M> 0 be given" but end by giving a condition on M. You want to show that, for any M, there exist $\delta$ such that if $|x- 1|<\delta$ then $\frac{x}{x- 1}> M$.

In order that $\frac{x}{x- 1}> M$, we must have either $x> M(x- 1)$ (if x> 1) or $x< M(x- 1)$ (if x< 1).

In the first case, x> 1, $x> Mx- M$ so $x- Mx= (1- M)x> -M$ and therefore $(M-1)x< M$. So $x< \frac{M}{M-1}$, and $x- 1= |x- 1|< \frac{M}{M-1}- 1=\frac{1}{M-1}$. That is, we can take $\delta$ to be less than $\frac{1}{M- 1}$.

I will leave the case x< 1 to you.

Answer 3

Note:

$\dfrac{x-1+1}{x-1} = 1+\dfrac{1}{x-1}.$

1) Consider $ x \rightarrow 1^+:$

Choose $M$, real, positive, and

$\delta >0$ with $\delta < 1/M.$

Then $|x-1|= x-1 \lt \delta$ implies

$M < 1+ 1/\delta < 1+\dfrac{1}{x-1} .$

2) Remains to consider $x \rightarrow 1^-.$