Does $f(x) = 1 = \frac{x-1}{x-1}$ have a hole at $x=1$?

There is no vertical asymptote. A function $f(x) = \frac{P(x)}{Q(x)}$, where $P$ and $Q$ are continous, has a vertical asymptote at $x_0$ if $Q(x_0)=0$ and $P(x)\neq 0$, which is not true in your case.

In your case, the function $f(x)=\frac{x-1}{x-1}$ simply has a hole.

As for why that hole appears:

Simple: It's because $1=\frac{x-1}{x-1}$ is not true for $x=1$. It is only true for $x\in\mathbb R\setminus \{1\}$.

You say in the beginning that "obviously", it is true that $a=\frac{a\cdot b}{b}$, which is an equality that is only true for $b\neq 0$.

In other words, your sentence:

Somehow, multiplying this extremely simple function by something equivalent to 1 makes it undefined at a particular point...

is not true, because you didn't multiply it by something equivalent to $1$. You multiplied it with something that equivalent to $1$ in most points and is undefined at a particular point.

Tags:

Calculus

Rational Functions

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