The question is the following: $\lim_{x \to 1^-} \frac{1}{1-x} = \infty$
I have done some rough work and a proof(that is probably incorrect), but I am not exactly sure what a correct proof should look like for this particular type of limit.
Rough work:
$W.T.S.: \forall M > 0, \exists \delta > 0 \hspace{4pt} s.t. \hspace{4pt} 0 < 1- x < \delta \Rightarrow \frac{1}{x-1} > M$
$1-x < \frac{1}{M}$
$\delta \leq \frac{1}{M}$
$Proof. Given \hspace{4pt} M > 0, choose \hspace{4pt} \delta = \frac{1}{M}. Assume \hspace{4pt} 0 < 1 - x < \delta.$
then $1 - x < \delta$
We have proven $\frac{1}{1-x} > M$, as needed.
Q.E.D.
The following is how I would prove it. Use it as a comparison. Do not take it as the absolute only way to write the proof. In mathematics, we allow personalization in proofs. I am in the camp that frowns upon using logic notation outside a logic setting. If you adopt this philosophy, then you may find yourself getting a better response when showing your proof to others (especially professors and teaching assistants!).
Proposition. $\lim_{x\to 1^-}\frac{1}{1-x}=+\infty$. That is to say, for each $M$ (positive integer; usually large), there exists $\delta>0$ such that for each real number $x$, if $1-\delta<x<1$, then $M<\frac{1}{1-x}$.
Proof. Fix an arbitrary $M$. Define $\delta:=\rule{30pt}{0.4pt}$. Fix real number $x$. Assume $1-\delta<x<1$. Observe $$\begin{align} 0<1-x&<\delta,\\ \frac{1}{1-x}&>\frac{1}{\delta}>0. \end{align}$$ Therefore if we choose $\delta:=\frac{1}{M}$, then we obtain $\frac{1}{1-x}>M$ as desired. $\square$