Thread: l'Hopital's Rule

I'm having a discussion (read: argument) with my calculus teacher about this.

Code:

lim      cos(x) - .5
          ---------
h->PI/6   PI/6 - x

I say that if you evaluate this limit it turns out undefined. She says that since the expression results in division by zero, l'Hopital's rule can be used.

Code:

lim      sin(x)
         ---------
h->PI/6    1

and this results in 0.5


Who is right? I think I am, but I want a general concensus here so I don't keep fighting a losing battle here.

Cshot, I'm not sure because you would end up with .366/0, and that isn't possible. also, yes, it should be x-> pi/6, probably just a typo.

Yes, your teacher is right though, hopefully not with this problem. L'Hopitals Rule can be used with these following inderterminate forms:

Code:

0/0
infinity/infinity
infiniti - infinity

i think there is more, i just can't remember them off the top of my head right now, and my calc notes are at school.

this particular limit does not classify as being able to use L'Hopital's Rule, this limit is undefined. I ran this through my calculator, and it gave me an answer of undefined (wonderful uses of the TI-89). If cos(x) - .5 as x -> pi/6 evaluated to zero, then i can see this working. as i hope this particular problem is not the one being discussed and does have typos.

My Mathmatical Theory (which you people are free to shoot down and you probably will):


I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.

Originally posted by golfinguy4
My Mathmatical Theory (which you people are free to shoot down and you probably will):


I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.

you can't take the ln of negative numbers, so the integral from -2 to 2 of 1/x dx would not equal zero. and inf/inf is indeterminate, because infinity is a concept. inf/inf could be 1, could be 5, etc.

Why do you mention ln? Yes, ln is the anti-derivative of 1/x. However, if you read the FToC carefully, you will see that the antidifferentiation rule is only guaranteed to work if f is continuous along the interval.

>> I think that inf/inf should equal 1. x/x=1 for non-zero integers.
Nah, you can't apply normal math operations with infinity. I think they're called indeterminants and normal math rules don't apply.



Hmm, alpha you may be right.
I looked it up here:
http://www.alltel.net/~okrebs/page191.html

It says:
If lim f(x) != 0 and lim g(x) = 0 then one of the following is true:
a) lim f(x)/g(x) = inf or lim f(x)/g(x) = -inf
b) lim f(x)/g(x) doesn't exist

Seems like if you approach pi/6 from the right, you'll approach negative infinity. If you approach pi/6 from the left, you'll approach positive infinity. Therefore, the limit doesn't exist for this case.

I think you are right. This is because you can only use lhr
when 0/0, inf/inf or -inf/inf.

You might want to show this to
your teacher

1/h as h->0 is certainly undefined. What follows gives
a similar expression

(cos(x) - .5) / (pi/6 -x) as x->pi/6 is equivalent to
(cos(x + h) - .5)/h as h->0 where x is fixed to be pi/6.
This is the same as
(cos(x + h) - cos(x) + (sqrt(3) - 1)/2) / h as h->0 which comes to
-sin(x) + (sqrt(3) - 1)/2h as h->0

I think that inf/inf should equal 1. x/x=1 for non-zero integers. Even the lim as h->inf of h/h=1. Therefore, since infinity is just a very very big number, shouldn't the same be true? This would make the integration of divergent functions very easy. For instance, with this idea, int from -2 to 2 of 1/x dx would equal 0.

If inf/inf equals 1 then x / (2x) as x->inf would be 1.

Wouldn't you just integrate 1/x and get ln |x| from -2 to 2 and
so then you have 0. integral(1/x) is not ln x.