Vertical Asymptote: Is f Defined at x=1?

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The discussion centers on whether a function can be defined at a point where there is a vertical asymptote. It is clarified that if x=1 is a vertical asymptote of y=f(x), then f is not defined at x=1, making the initial statement false. Counterexamples are provided, such as f(x)=1/(x-1) which is undefined at x=1, while other functions can be defined at that point but still have an asymptote elsewhere. The participants agree that a function can be defined at a point while having an asymptote at a different x-value, but the specific case of x=1 as an asymptote contradicts the function being defined there. Ultimately, the conclusion is that a vertical asymptote implies the function is not defined at that point.
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Homework Statement



True False

If the line x=1 is a vertical asymptote of y = f(x), then f is not defined at 1.

Homework Equations



none

The Attempt at a Solution



I originally believed this was true, but on finding it was false it sought a counter example:

if for example f(x) = 1/x if x != 0
5 if x = 0

Then the function is defined, but the asymptote still is at x=1, correct?

This is very basic - I just want to make sure I understand it thoroughly. Thanks.
 
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dkotschessaa said:

Homework Statement



True False

If the line x=1 is a vertical asymptote of y = f(x), then f is not defined at 1.

Homework Equations



none

The Attempt at a Solution



I originally believed this was true, but on finding it was false it sought a counter example:

if for example f(x) = 1/x if x != 0
5 if x = 0

Then the function is defined, but the asymptote still is at x=1, correct?

This is very basic - I just want to make sure I understand it thoroughly. Thanks.

The vertical asymptote for that example is at x=0.

So yes the function is defined at x=1 since if we plug x=1 into the equation, we get 1. The asymptote isn't at x=1 though.
 
Thanks Mentallic!

So you're right. Not a great example. So how about [1/(x-1)] - 1 with f(1) = 5 (or some number)

Point being I guess, that a function can still be defined where there is an asymptote.
 
p.s. Posting a limit problem over in the calc forum, if you're feeling especially helpful today. This question was actually from my calc book.
 
dkotschessaa said:
Thanks Mentallic!

So you're right. Not a great example. So how about [1/(x-1)] - 1 with f(1) = 5 (or some number)
If you define the function to be defined at x=1, then that's what it's going to be. But the function f(x)=1/(x-1) alone is not defined at x=1.

dkotschessaa said:
Point being I guess, that a function can still be defined where there is an asymptote.
As you've done, yes, but the question was implying there are conditions such as the obvious - you can't define it to be defined at that x value :wink:
 
Well, I was just trying to come up with any example that would serve as a situation where 1) - there is an asymptote at some x and
2) the function is defined at x

I'm sure there are other examples.

Thanks again!
 
dkotschessaa said:
Well, I was just trying to come up with any example that would serve as a situation where 1) - there is an asymptote at some x and
2) the function is defined at x

I'm sure there are other examples.

Thanks again!

Well yes, under a certain set of conditions. The answer to the problem is no however.
 

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