Asemptotes problem

Aug 2007
21
0
Hello,
Ive started a problem where I have to fine the asymptote of the function \(\displaystyle y = (x^3)/(x^2-1)\)

I know the asemptotes are \(\displaystyle x=-1, x=+1\)

But with using limits I get the wrong answer. (being the opposite to what I need)

for example lim x->-1(on the positive side) i should get (+ve infinity.) But when I do the math (using a calculator) I get -ve infinity.

Cheers for any help
~Regards
Name101
 

mr fantastic

MHF Hall of Fame
Dec 2007
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Hello,
Ive started a problem where I have to fine the asymptote of the function \(\displaystyle y = (x^3)/(x^2-1)\)

I know the asemptotes are \(\displaystyle x=-1, x=+1\)

But with using limits I get the wrong answer. (being the opposite to what I need)

for example lim x->-1(on the positive side) i should get (+ve infinity.) But when I do the math (using a calculator) I get -ve infinity.

Cheers for any help
~Regards
Name101
Vertical asymptotes: Vertical lines that pass through the values of x such that x^2 - 1 = 0.

And since \(\displaystyle \frac{x^3}{x^2 - 1} = x + \frac{x}{x^2 - 1}\) it should be evident that there is also a diagonal asymptote with equation y = x.
 
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Jan 2010
43
7
1: \(\displaystyle \lim_{x \to -1^-} [ \frac{x^3}{x^2 - 1}] = \frac{minus}{plus} = minus\)

\(\displaystyle \frac{(-1)^3}{(-1.0001)^2 - 1} = \frac{-1}{0.00020001} =\) negative!


2: \(\displaystyle \lim_{x \to -1^+} [ \frac{x^3}{x^2 - 1}] = \frac{minus}{minus} = plus!\)

3: \(\displaystyle \lim_{x \to 1^-} [ \frac{x^3}{x^2 - 1}] = \frac{plus}{minus} = minus!\)

4: \(\displaystyle \lim_{x \to 1^+} [ \frac{x^3}{x^2 - 1}] = \frac{plus}{plus} = plus!\)
 
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Aug 2007
21
0
Mr Fantastic~ "it should be evident that there is also a diagonal asymptote with equation y = x."

Hmmm if there is a asymptote at y=x, wouldn't the function discontinue at x=0 and y = 0?

Sponsoredwalk:
If i was talking the the limit of \(\displaystyle \lim_{x \to -1^-} [ \frac{x^3}{x^2 - 1}]\)
would it be
\(\displaystyle \frac{(-0.99999)^3}{(-0.99999)^2 - 1} = \frac{-1}{-1.9999*10^{-4}}\) = Minus/Minus = positive??

Or is that where i'm going wrong?

Thanks for the help so far.
 
Jan 2010
43
7
You don't need to use scientific notation in evaluating this limit.

All you're looking for is the signs on top & on the bottom.

Now, I would guess you are not fully aware of what I wrote because in the limit you've provided you've mixed it up.

\(\displaystyle \lim_{x \to -1^-} [ \frac{x^3}{x^2 - 1}]\)

There is a subtlety here, in \(\displaystyle \lim_{x \to -1^-}\) do you see the minus on top of the (-1)?

This means that you use the number line & go from -∞ towards -1, i.e. if you substitute in -2 it will work here because all you're interested in is the sign.

The minus there means that you approach whatever the limit is from that side.

\(\displaystyle \lim_{x \to -1^-}\) approach (-1) from the negative side & increase towards the positive side.

\(\displaystyle \lim_{x \to -1^+}\) approach (-1) from the positive side & go down towards the negative side.

Also, about the asymptote, you see that Mr. Fantastic gave you
\(\displaystyle \frac{x^3}{x^2 - 1} = x + \frac{x}{x^2 - 1}\)
well he just used long division to reduce the original equation into an equivalent equation but it cannot be used at (-1) or 1 because the original function is undefined there.

This should be instantly recognisable because the degree in the numerator is one higher than that of the denominator.
When you see this type of function you use long division to reduce it to an equivalent equation & the first term that is your asymptote, (in this case x!).

The function is well defined at 0, it is equal to zero!

 
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mr fantastic

MHF Hall of Fame
Dec 2007
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Mr Fantastic~ "it should be evident that there is also a diagonal asymptote with equation y = x."

Hmmm if there is a asymptote at y=x, wouldn't the function discontinue at x=0 and y = 0?

[snip]
Have you been taught about other types of asymptotes apart from vertical asymptotes?
 
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Aug 2007
21
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Have you been taught about other types of asymptotes apart from vertical asymptotes?
No, I think we may have touched on the subject but not for long.

There is a subtlety here, in \(\displaystyle \lim_{x \to -1^-}\) do you see the minus on top of the (-1)?
I did see the little -, but this is where I am confused. With the little minus there does that mean that x is slightly smaller than -1?

Thanks for the help so far :D
 
Jan 2010
43
7
Yeah it does mean that you are slightly smaller than (-1), like say (-2) or anything.

I have a load of links to video lectures that you might appreciate:

http://www.uccs.edu/~math/vidarchive.html (requires registration but is free)
Math 112
MATH 113
Math 115 New
http://www.sci.uidaho.edu/polya/math...=1&sec=1&sub=0
Free downloadable Maths Video Lecture courses
http://www.math.ncsu.edu/calculus/we...1lectures.html
Department of Mathematics - NC State University
http://courses.ncsu.edu/ma107/common...7Lectures.html
http://webcast.berkeley.edu/course_d...sid=1906978460
www.khanacademy.org
Just Math Tutoring
Calculus-Help.com: Survive calculus class! - Tutorials for the Calculus Phobe
5min - Find the best how to, instructional and DIY videos ? Life Videopedia (type in thinkwell & use the calc videos!).

all free and usable, I think most are calc-related & should do everything to clear up math woes :p

I think you need to learn about limits & asymptotes from a proper source, I'd suggest Khan Academy as a starter then the first link, uccs as a kind of proper college teacher explaining these things to get both views.
 
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