1. ## Vertical Asymptotes

f(x) = (1 - x) / [(1 - x)(5 + x)]

My professor argues with us that the value 1 is considered a vertical asymptote.

Is he correct?

2. ## Re: Vertical Asymptotes

no vertical asymptote ... there is a point discontinuity (a hole) at x = 1

see attached TI-84 graph ... tick mark on the x-axis is x = 1

3. ## Re: Vertical Asymptotes

i know that there is a hole at x = 1, but my professor consider this point of discontinuity as a vertical asymptote.

How to convince him that it is not?

Arguing with professors is dangerous cuz they are always right!

4. ## Re: Vertical Asymptotes

Sounds like it's time to stand up, show your evidence, and argue your point ... do you have face-to-face contact with this "professor", or is he/she some online hack? Don't be intimidated. What can he/she do to you? ... take away your birthday?

5. ## Re: Vertical Asymptotes

He is with me face to face in class. I feel that he made me wrong cuz he does not want to show students that he got a mistake, but if x = 1, will never be a vertical asymptote, i can discuss this issue with him again in his office. I just want to make sure he is wrong cuz if he is right, i am confused about this answer.

6. ## Re: Vertical Asymptotes

Originally Posted by joshuaa
He is with me face to face in class. I feel that he made me wrong cuz he does not want to show students that he got a mistake, but if x = 1, will never be a vertical asymptote, i can discuss this issue with him again in his office. I just want to make sure he is wrong cuz if he is right, i am confused about this answer.

-Dan

7. ## Re: Vertical Asymptotes

Thanks alot guys. now, i can show him my muscles in this issue.

8. ## Re: Vertical Asymptotes

This is the problem we discussed in class.

Determine all horizontal and vertical asymptotes. For each vertical asymptote, determine whether f(x) goes to infinity or f(x) goes to negative infinity on either side of the asymptote.

f(x) = (1 - x) / (x^2 + x - 2)

9. ## Re: Vertical Asymptotes

fyi, this is not the same function you originally posted $\rightarrow$
f(x) = (1 - x) / [(1 - x)(5 + x)]

f(x) = (1 - x) / (x^2 + x - 2)
$f(x) = \dfrac{1-x}{(x+2)(x-1)} = - \dfrac{x-1}{(x+2)(x-1)}$

this function still has a point discontinuity at x = 1

10. ## Re: Vertical Asymptotes

What definition does your teacher or text give for "vertical asymptote"?

11. ## Re: Vertical Asymptotes

But it has the same concept.

I know that x = -2 is a vertical asymptote.

But would you consider x = 1 as a vertical asymptote?

12. ## Re: Vertical Asymptotes

Originally Posted by joshuaa
But it has the same concept.

I know that x = -2 is a vertical asymptote.

But would you consider x = 1 as a vertical asymptote?
a vertical asymptote is an infinite discontinuity, not a point discontinuity

for $f(x) = -\dfrac{x-1}{(x+2)(x-1)}$ ...

x = -2 is a vertical asymptote , point discontinuity at x = 1, horizontal asymptote is y = 0

13. ## Re: Vertical Asymptotes

May be this is the definition.

Examine lim 1/x as x goes to 0

Of course, we can draw a graph and compute a table of function values easily, by hand. While we say that the limits lim 1/x as x goes to 0+ and lim 1/x as x goes to 0- do not exist, they do so for different reasons. Specifically, as x goes 0+, 1/x increases without bound, while as x goes 0-, 1/x decreases without bound. To indicate this, we write

(1) lim 1/x as x goes 0+ = Infinity

and

(2) lim 1/x as x goes 0- = -Infinity

Graphically, this says that the graph of y = 1/x approaches the vertical line x = 0, as x goes to 0. When this occurs, we say that the line x = 0 is a vertical asymptote. It is important to note that while the limits (1) and (2) do not exist, we say that they equal Infinity and -Infinity, respectively, only to be specific as to why they do not exist. Finally, in view of the one-sided limits (1) and (2), we say as before that

lim 1/x as x goes 0 does not exist

14. ## Re: Vertical Asymptotes

Originally Posted by skeeter
a vertical asymptote is an infinite discontinuity, not a point discontinuity

for $f(x) = -\dfrac{x-1}{(x+2)(x-1)}$ ...

x = -2 is a vertical asymptote , point discontinuity at x = 1, horizontal asymptote is y = 0
Thanks a Lot skeeter. This is exactly what I want to reach to.

15. ## Re: Vertical Asymptotes

Originally Posted by joshuaa
I know that x = -2 is a vertical asymptote.
But would you consider x = 1 as a vertical asymptote?
$f(x) = \frac{{(1 - x)}}{{(1 - x)(5 + x)}} = \frac{1}{{(5 + x)}},\text{ if }x \ne 1$ then $\displaystyle{\lim _{x \to 1}}f(x) = \frac{1}{6}$

If your instructor sees this and then insists he/she is correct, then take the matter to the head of school.

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