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While learning about asymptotes and holes in rational functions in Precalculus, I came across a problem that shouldn't happen, but I don't understand. Something just doesn't add up in my head... Here we go:

Let f(x)=1. Since $\frac aa=1$ and $a*1=a$, obviously $a={a*b \over b}$. Let $b=(x-1)$. We can multiply f(x) by 1, technically, by saying that: $$f(x)=1={x-1 \over x-1}$$ However, a bit of precalc knowledge shows us that ${x-1 \over x-1}$ has a vertical asymptote at x=1, and an x-intercept at x=1 as well. This means that the graph has a hole, or is undefined at, x=1. (Because plugging in 1 for x yields a denominator of 0, which is undefined.) However, this is absurd because the graph of $f(x)=1$ obviously has no holes, anywhere. Somehow, multiplying this extremely simple function by something equivalent to 1 makes it undefined at a particular point...

Whaaaaa...?

EDIT: I realize that the above function does not have a vertical asymptote. However, my statement was referring to the fact that it appears to have both a vertical asymptote and an x-intercept at the same x-value, which means that it has neither of those things, just a hole.

Najib Idrissi
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vasilescur
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    There is no vertical asymptote at $x=1$ for $\tfrac{x-1}{x-1}$, but there is a 'hole'... – StackTD Feb 24 '16 at 13:10
  • Related. – Cameron Buie Feb 24 '16 at 13:25
  • Wow there is such a confusion in your head... first "hole" is not a term used in math so I think you are talking about discontinuities, second "a denominator of $0$" is not undefined unless even the numerator is $0$ and this is the case, by the way taking limit you get that $f(1)=1$ -a third order discontinuity- as you would expect so you can redefine your function as $f(x)=1$ doing actually nothing useful. – AlienRem Feb 24 '16 at 13:40
  • @RenatoFaraone Dude. The OP is clearly a beginner, and starting your answer with "Wow there is such a confusion in your head" isn't really helping. I'm sure there are things you don't understand. How would you like it if your professor, instead of explaining them to you, would start immediatelly shouting at you how stupid you are for not understanding them? Further more, there is no discontinuities in the function $\frac{x-1}{x-1}$, so before shouting at OP for using the wrong terms, check your own terms. – 5xum Feb 24 '16 at 13:43
  • @5xum sorry if I sounded aggressive but it was not in my intention, I was simply worried. By the way in my book it is clearly said that those kind of function has "discontinuities of the third kind" or "eliminable discontinuities". – AlienRem Feb 24 '16 at 13:46
  • You are showing your misunderstanding in the title. You have $\frac{x- 1}{x- 1}$ is NOT equal to 1! It is equal to 1 for x not equal to 1 and is undefined at x= 1. That is where the "hole" comes in. – user247327 Aug 20 '19 at 17:55

2 Answers2

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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.

CAGT
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5xum
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    Thank you, now I realize where I went wrong. – vasilescur Feb 24 '16 at 13:15
  • I uh, can't actually... The button to accept is simply not there (on mobile app, iPad), and I "do not have enough reputation to change the publicly displayed score"... Edit: there we go, worked now – vasilescur Feb 24 '16 at 13:42
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When you make divisions like $\frac{f(x)}{p(x)}$, always make the claim that p(x) should be different from 0.

Socre
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