Let $f \colon \mathbb{R} \setminus \{ 1 \} \to \mathbb{R}$ be the function defined by $$ f(x) \colon= \frac{x}{x-1} \ \mbox{ for all } x \in \mathbb{R} \setminus \{ 1 \}. $$ Then how to prove rigorously the following limit statement? $$ \lim_{x \to 1 - } f(x) = -\infty. $$
This is Prob. 5 (b), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.
Here is my attempt at a rigorous proof of the above statement.
Let us first restrict our $x$ such that $$ 0 < x < 1. \tag{0} $$ Then $x-1 < 0$ and so $$ \frac{x}{x-1} < 0. \tag{1} $$
Now let us take any real number $\alpha$.
CASE 1: If $\alpha \geq 0$, then we find from (1) that $f(x) < \alpha$ for all $x \in \mathbb{R}$ such that $0 < x < 1$.
So if we choose a real number $\delta$ such that $0 < \delta < 1$, then $-1 < -\delta$ and hence $0 < 1-\delta$, and thus for all $x \in \mathbb{R}\setminus \{ 1 \}$ such that $ 1 - \delta < x < 1$, we have $0 < x < 1$ which implies that $f(x) < \alpha$.
CASE 2: Now let us take our $\alpha < 0$.
Then for all $x \in \mathbb{R}$ for which (0) above holds, we see that $f(x) < \alpha$ holds if and only if $$ \frac{x}{x-1} < \alpha, $$ and this holds if and only if $$ x > (x-1) \alpha, $$ [ because $x-1 < 0$, by virtue of (0) ] or $$ x > x \alpha - \alpha \tag{2} $$
Now (2) holds if and only if $$ \alpha > x \alpha - x = x(\alpha - 1), $$ and this holds if and only if $$ \frac{\alpha}{\alpha - 1} < x, \tag{3} $$ because $\alpha - 1 < 0$. [ Note that here we have assumed that $\alpha < 0$. ]
Now the left-hand side of (3) can be rewritten as $$ 1 - \frac{1}{1-\alpha} < x. \tag{3} $$
Now let us choose any real number $\delta$ in such a manner that $$ 0 < \delta \leq \frac{1}{1-\alpha}. $$ Then $$ - \frac{1}{1-\alpha} \leq - \delta, $$ and hence $$ 1 - \frac{1}{1- \alpha} \leq 1 - \delta. \tag{4} $$
So for all $x \in \mathbb{R}$, if $$ 1 - \delta < x < 1, $$ then from (4) we can conclude that for all those $x$ we also have $$ 1 - \frac{1}{1- \alpha} < x < 1, $$ and so from (0) and (3) we obtain $$ f(x) < \alpha. $$
Thus in either case we have shown that, for every real number $\alpha$, we can find a real number $\delta > 0$ so that $f(x) < \alpha$ for all $x \in \mathbb{R}$ which satisfy $$ 1-\delta < x < 1. $$
Hence $$ \lim_{x \to 1-} \frac{x}{x-1} = - \infty, $$ as required.
Is this proof correct and clear enough? If not, then where are the deficiencies?
It seems correct, but it's too long and verbose.
You have to prove that, for every $\alpha>0$, there exists $\delta>0$ such that, for $1-\delta<x<1$, $$ \frac{x}{x-1}<-\alpha $$ This can be rewritten as $$ \frac{x(1+\alpha)-\alpha}{x-1}<0 $$ which is satisfied for $$ \frac{\alpha}{1+\alpha}<x<1 $$ Note that $$ \frac{\alpha}{1+\alpha}=1-\frac{1}{1+\alpha} $$ and take $\delta=1/(1+\alpha)$.
You don't even discuss the case $\alpha\le0$; if you're pedantic, just take $\delta=1/2$ for every $\alpha\le 0$ (the value of $\delta$ corresponding to $\alpha=1$.