Horizontal Asymptote

[Ergot]

Where is it when y = 3x / (x+3)^2 ?

[dark_drake]

y = 0

At least if I remember correctly.

[Ergot]

Explain ?

[dark_drake]

The limit as x approaches +infinity and -infinity is 0.

[Sidoh]

Explain ?

Actually, there's a really easy way to do this problem (without graphing or anything).

Since the total powers of x (2) are greater in the denomonater than they are in the numerator, f(x) will approach 0 as x approaches infinity.  This is always the case with any function, no matter how big a constant on either may be.  Conversely, if the powers of x are larger in the numerator than in the demononator, f(x) will approach infinity.

[dark_drake]

Explain ?

Since the total powers of x (2) are greater in the denomonater than they are in the numerator, f(x) will approach 0 as x approaches infinity.  This is always the case with any function, no matter how big a constant on either may be.  Conversely, if the powers of x are larger in the numerator than in the demononator, f(x) will approach infinity.

This is true; however, learning the limit thing is useful for when the powers are equal.  As always, I reccomend learning the basics and using shortcuts after growing comfortable with the basics.

[Ergot]

Thanks... I have a quiz on this tomorrow, hopefully he'll leave those types out...
drake... explain the basic to me? AFAIK, this will get me pretty far until I learn more in calc or w/e.

[dark_drake]

Thanks... I have a quiz on this tomorrow, hopefully he'll leave those types out...
drake... explain the basic to me? AFAIK, this will get me pretty far until I learn more in calc or w/e.

Just the limit as x approaches plus and minus infinity. How you do that, at least how I learned, is about like this:


Basically, you can't evaluate the limit of the function as x approaches infinity because when you plug infinity in, you get infnity/infinity, which is undefined.  So, you have to divide the numerator and denominator by the largest power of x.

Edit: This only works if the powers in the denominator are >= those in the numerator.  If the numerator's are greater, you're dealing with slant asymptotes.

[Ergot]

I SEE !!!

[rabbit]

Or when it's undefined (IE: -3).

[Ergot]

Or when it's undefined (IE: -3).

That would be x = -3 or the vertical asymptote... that one is easy.

[rabbit]

Fuck asymptotes, I'm done with high school!

[MyndFyre]

Thanks... I have a quiz on this tomorrow, hopefully he'll leave those types out...
drake... explain the basic to me? AFAIK, this will get me pretty far until I learn more in calc or w/e.

Just the limit as x approaches plus and minus infinity. How you do that, at least how I learned, is about like this:


Basically, you can't evaluate the limit of the function as x approaches infinity because when you plug infinity in, you get infnity/infinity, which is undefined.  So, you have to divide the numerator and denominator by the largest power of x.

Edit: This only works if the powers in the denominator are >= those in the numerator.  If the numerator's are greater, you're dealing with slant asymptotes.

Easier way to do that limit: L'Hospital's Rule. 



Really handy, because it can be applied several times.  For example, you want to find the limit of the following function:


So, as a good, easy mneumonic device:

1.) If the exponent of the numerator is greater, the function will have a limit of +infinity or -infinity.
2.) If the exponent of the denominator is greater, the function will have a limit of 0.  Series defined by this function will be convergent.
3.) If the exponents of the numerator and denominator are the same, the limit will be a constant of the numerator's greatest-power coefficient times the factorial of the power, divided by the denominator's greatest-power coefficient times the factorial of the power (in the example it was 6*5!/12*5!).  I believe that the series will be convergent, but I can't remember for sure.

Also, limits don't need to be taken to infinity.  They can be found at any point along the function.

[Ergot]

Uhh... I don't understand that I'm only in Algebra II.

[deadly7]

What the hell do factorials have to do with this?  You have weird ways of doing things.. graphing takes about two seconds with a calculator and is way easier.

[d&q]

Uhh... I don't understand that I'm only in Algebra II.

Myndfyre's way is the way my Algebra II teacher taught us. You don't really need to understand all the calculus stuff.

[MyndFyre]

What the hell do factorials have to do with this?  You have weird ways of doing things.. graphing takes about two seconds with a calculator and is way easier.

Yeah, but if you can't use a graphing calculator or need to justify it on the test, my mneumonic devices give you way more guidance.

Factorials come into play because of derivation when you're dealing with polynomials (I should have included the "polynomial" qualifier in my earlier statement, because this won't work for things like logarithms and inverse fractions).  If you know d/dx 4x⁴ = 16x³, and that you keep applying this until the power of x is 0, you'll get 96.  That's the coefficient 4 times 4!. 

Myndfyre's way is the way my Algebra II teacher taught us. You don't really need to understand all the calculus stuff.

Well, you need to understand the calculus stuff to understand why Rule 3 applies (it's because you're repeatedly taking the derivative of the highest-power component of the polynomial, since everything else drops to 0). 

[d&q]

All you need to know is limits I believe.

[Nate]

Chill peoples. 

Horizontal Asymptope-
Numerator = 0

Vertical Asymptope-
Denomonator = 0

[MyndFyre]

Chill peoples. 

Horizontal Asymptope-
Numerator = 0

Vertical Asymptope-
Denomonator = 0

That is incomplete.  The function f(x) = 3 / (2x + 6) is horizontally asymptotic at y = 0 even though the numerator is never 0.  Your definition does not incorporate all possibilities.

[Rule]

Also, limits don't need to be taken to infinity.  They can be found at any point along the function.

Not quite true.  The numerator and the denominator have to both be approaching infinity or zero for l'hopital's rule to work. 

For example

lim x--> 2    [3x+2] / (3x)  = 8/6, but lim x-->2 of d/dx(3x+2)/ d/dx(3x) = 3/3 = 1 != 8/6.

Challenge:  Prove L'hopital's rule.
Trivia: L'Hopital bought L'Hopital's rule from the Bernoulli family .

(Nate):  Also, consider 0/0.  And, if y =0 and the denominator is nonzero real, there is an horizontal asymptote at y=0.  But there can be horizontal asymptotes at y={any real number}.  Same goes for vertical asymptotes.
An horizontal asymptote is when y --> or = constant as x--> infinity, or x--> -infinity.  A vertical asymptote is when
x --> or = constant as y--> infinity or -infinity. 

[Rule]

Ah, I'll do it; the proof is kind of cool, and might help people understand limits better, etc..

(Note, the proof was mostly done from scratch, so there may be errors):
The Mean Value Theorem:
If a function w is continuous, real valued, and differentiable on the real interval [a,b], there exists a c in [a,b] such that

w'(c) = [w(b)-w(a)] / [b-a].  Essentially the derivative at some point along the curve is equal to the slope of the secant line from a to b (this is a bit intuitive).


Apply the mean value theorem to
r(x) = f(x)[g(b)-g(a)] - g(a)[f(b)-f(a)]

You will find that
f'(c) / g'(c)  = [f(b)-f(a)] / [g(b)-g(a)]

Now imagine f(a) and g(a) --> 0. 
f'(c) / g'(c) = f(b)/g(b) .   Now since c is in [a,b], if we let b-->a, then c must also approach a.
So, f'(b)/g'(b) = f(b)/g(b) so long as f(b) , g(b) --> 0 as b --> a.

This is also why L'Hopital's rule works if the numerator and denominator both approach something infinite (e.g. numerator could --> infinity, denominator could approach --> -infinity):

Assume w(b) / m(b) = infinite value / infinite value as b-->a

[1/w(b)]  / [1/m(b)] =  0/0 as b--> a  = d/db [ 1/w(b) ] /  d/db[1/m(b) ]
Using the chain rule,
 m(b)/w(b)  = [-w'(b)/w(b)²] / [-(m'(b)/m(b)²]
= w'(b)*m(b)² / [w(b)²*m'(b)]
or,
w(b)/m(b) = w'(b)/m'(b)  as b--> a.

That was fun  .   I was a bit scared for a minute because it didn't look like the infinite case would work out at first.

[Rule]

3.) If the exponents of the numerator and denominator are the same, the limit will be a constant of the numerator's greatest-power coefficient times the factorial of the power, divided by the denominator's greatest-power coefficient times the factorial of the power (in the example it was 6*5!/12*5!).  I believe that the series will be convergent, but I can't remember for sure.

I just noticed this.  I'm not sure what you mean "series will be convergent.."?

Rule 3) can be simplified. 
"If the [degree of the polynomial in] the numerator and denominator are the same, the limit will be" equal to the quotient of the coefficient of the highest power in the numerator to the coefficient highest power in the denominator.   

Using L'Hopital's rule (in this case only applicable if the limits are taken at infinity):       
e.g.  [a*x^n + ... lower order polynomial]/[b*x^n + ... lower order polynomial]
lim x--> infinity
= [a*n! + (0.. the lower order polynomial has dissapeared through differentiation) ] / (b*n!)
the factorials cancel, and you are left with a/b

Preferable (no calculus required):
If the limits are approaching infinity, divide top and bottom by x^n
lim x --> infinity  [a*xⁿ/xⁿ + ... lower order polynomial/xⁿ]/[b*xⁿ/xⁿ + .. lower order polynomial / (xⁿ)]

= a/b   as  the x^n approaches infinity faster than a polynomial of degree less than n