# l'Hopital's Rule

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• 01-16-2003
ygfperson
l'Hopital's Rule

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.
• 01-16-2003
Polymorphic OOP
your teacher is corect, however you are confused.

The limit has the potential of existing not because the denominator is 0, but because BOTH the numerator AND the denominator are 0.

In other circumsatnces (IE without trigonometric functions) you can do things like factoring and cancelation (logic being that 0/0 means both the numerator and the denominator have a common factor) or use other methods.
• 01-16-2003
Cshot
I think you're both wrong.

Shouldn't it be x->pi/6 instead of h->pi/6 :D

Anyways, shouldn't the answer be infinity since the numerator isn't zero at x=pi/6?
• 01-16-2003
alpha
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.
• 01-16-2003
Cshot
>> Cshot, I'm not sure because you would end up with .366/0, and that isn't possible.
You're right that .366/0 is undefined. But this is a limit problem. As x->pi/6, the limit goes to infinity.

And yea, I was just joking about the h->pi/6 part :)
• 01-16-2003
alpha
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.
• 01-16-2003
Cshot
I agree with you alpha that this isn't a l'hopital problem. Just solve it like any other limit problem. The numerator goes to a constant while the denominator approaches 0, therefore the entire limit approaches infinity.

If the problem were to say what is (cos(x)-0.5)/(pi/6 - x) at x=pi/6, then the answer would be undefined.
• 01-16-2003
Polymorphic OOP
ha, you're right, i didn't notice the numerator wasn't 0. You'd think that after I specifically said the numerator had to be 0 that would have checked if it was in this example :p

cos( pi/3 ) is 1/2 NOT cos( pi/6 )
• 01-16-2003
alpha
Cshot, it still doesn't make sense. The calculator does do limits and it says the answer is undefined. if the denominator approached 1, then i could see how it approaches a constant; but it approaches 0, and once pi/6 is plugged in just like any limit problem, it is evaluated with the values, and .366/0 is undefined.

I'm still in high school Calc BC, so maybe I'm missing something here...but I don't think I am.

edit: Actually, come to think of it, if the limit was as x approaches infinity, then the limit would evaluate to infinity...
• 01-16-2003
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.
• 01-16-2003
alpha
Quote:

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.
• 01-16-2003
golfinguy4
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.
• 01-16-2003
Cshot
>> 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.
• 01-16-2003
Nick
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

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
• 01-16-2003
Nick
Quote:

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.
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