Is this argument for uniform continuity good?

Question

Consider the function $f:(-\infty, 1)\to \mathbb{R}$, $f(x)=\frac{1}{x-1}$. I would like to prove that this function is uniformly continuous.
My argument is as follows: the function is continous because it is the division of two continuous functions. Now, $\displaystyle \lim_{x\to -\infty}f(x)=0$, so we are done.
I want to make sure that this is a correct way of proving uniform continuity. I know that this works for functions defined on an interval that contains $\infty$, but I want to make sure that this also works for $-\infty$.
Furthermore, what if our function's domain is no longer an interval, but a union of intervals that contains $\infty$ or $-\infty$? Does the conclusion still hold?

Answer 1

No, this is not true. The problem here is not $-\infty$, the problem happens near the point $1$. For example, $1-\frac{1}{n}$ is a Cauchy sequence of elements in $(-\infty,1)$. However, $f(1-\frac{1}{n})=-n$ is clearly not a Cauchy sequence. So $f$ is not uniformly continuous. (otherwise it would map Cauchy sequences to Cauchy sequences)

Another way to see this: in general, if $a,b$ are real numbers (not $\infty$ or $-\infty$) and a function $f$ is uniformly continuous at $(a,b)$ then it can be extended to a continuous function $[a,b]$. However, your function can't be continuously extended to the point $x=1$, as the limit $\lim\limits_{x\to 1^-} f(x)$ doesn't exist.