I need show that $\lim_{x\longrightarrow 1}\dfrac{1}{x-1}$ does no exist. My intent is the following, prove that $\lim_{x\longrightarrow 1^+}\dfrac{1}{x-1}=\infty$ and $\lim_{x\longrightarrow 1^-}\dfrac{1}{x-1}=-\infty$.
Let $\alpha\in \mathbb{R}$, then if $\delta=1/\alpha$ for every $x\in (1,\infty)$ where $0 < x-1< \delta$, then $\dfrac{1}{x-1}>\dfrac{1}{\delta}=\alpha$, so that $\dfrac{1}{x-1}>\alpha$, hence $\lim_{x\longrightarrow 1^+}\dfrac{1}{x-1}=\infty$
Let $\beta\in \mathbb{R}$, then if $\delta=1/\beta$ for every $x\in [0,1)$ where $0 < 1-x< \delta$, then $\dfrac{1}{1-x}>\dfrac{1}{\delta}=\beta$, so that $\dfrac{1}{x-1}<-\beta$, hence $\lim_{x\longrightarrow 1^-}\dfrac{1}{x-1}=-\infty$.
I have an error or you think it's okay.
It's all good, but note that for this particular purpose you don't need to compute limits, all you need is to observe that $$\cases{\frac{1}{x-1}<-1&for $x\in (0,1), $\\\frac{1}{x-1}>1&for $x\in (1,2). $}$$