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Limits that gives 0/0 - Page 3

Blogs > Raithed
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fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
May 27 2008 03:12 GMT
#41
On May 27 2008 10:37 Raithed wrote:
Waaaah, you people are too smart for me. Dx

how about you give us an example that can't be done with factoring
Do you really want chat rooms?
qet
Profile Joined May 2007
Australia244 Posts
May 27 2008 06:22 GMT
#42
On May 27 2008 12:12 fight_or_flight wrote:
Show nested quote +
On May 27 2008 10:37 Raithed wrote:
Waaaah, you people are too smart for me. Dx

how about you give us an example that can't be done with factoring



try: lim sin(x) / x
.... x->0

can't be done with factoring
tiffany
Profile Joined November 2003
3664 Posts
May 27 2008 07:40 GMT
#43
special case, just remember the limit is 1 =]
wanderer
Profile Blog Joined May 2007
United States641 Posts
Last Edited: 2008-05-27 07:53:38
May 27 2008 07:48 GMT
#44
god damnit i was beaten to the l'hospital flame.

i didnt see anyone tell the story about it though so here goes:

once upon a time, there were two brothers. they were called the Bernoulli Brothers. you've probably heard of them in science class for the Bernoulli Effect. well something you don't learn in class goes like this:

one day a rich aristocrat named L'Hospital was riding on the bandwagon as new discoveries in mathematics were being revealed. he knew that the people who discovered these things would be remembered for a long, long time; so he wanted in on it. when one of the Bernoulli brothers (i forget which one) discovered this rule, he was given a bunch of money by monsieur L'Hospital to name the rule after him, thus saving his name to posterity. Besides, there was already a Bernoulli Principle, so why not?

[/story]

lim x->5 [ x-5 / 4x-20 ]
=>
d [ x-5 ] / dx = 1
d [ 4x-20 ] / dx = 4
(this is L'Hospital's rule in action)
=>
lim x->5 [ 1 / 4 ] = 1/4


Important note:
you CAN NOT do this whenever you want. you can only do this when the limit, as it is worded, goes to 0/0 or (+/- oo) / (+/- oo).

the bernoulli brothers had no idea why this was true and couldn't prove it -- it just worked. weird coincidence. thats why its called "L'Hospital's Rule" instead of "L'Hospital's Theorem/Law".


[/history + math help]




HOW TO DO ANY L'HOSPITAL'S RULE PROBLEM:

[image loading]

courtesy of qet, although the name isn't spelled correctly.
Fuck you, I have a degree in mathematics and I speak 12 languages. (I called the World Cup final in 2008 btw)
qet
Profile Joined May 2007
Australia244 Posts
Last Edited: 2008-05-27 07:58:23
May 27 2008 07:54 GMT
#45
On May 27 2008 16:48 wanderer wrote:
courtesy of qet, although the name isn't spelled correctly.


according to wikipedia:
http://en.wikipedia.org/wiki/L'Hôpital's_rule
can be spelt either "l'Hôpital" or "l'Hospital" but i was taught no 's'.


jgad
Profile Blog Joined March 2008
Canada899 Posts
May 27 2008 07:56 GMT
#46
On May 27 2008 07:29 micronesia wrote:
Show nested quote +
On May 27 2008 05:45 L wrote:
Having 1 method to cover all problems of a set reduces the amount of study time you need to cover that area and allows you to progress more quickly on tests.

Depending on topic, this can be very costly. There are times when you need more than one method of solving a problem, even if you can theoretically solve it with just one method.


I agree. I teach at university level and one of the most important things, I feel, to get across to students is the ability to look at a problem from more than one angle. The idea that one can just store a rule like L'Hopital's and use it on a certain type of problem is the sort of thing that gets people stuck. In the worst case I find it produces the sort of people who understand that the rule will solve a certain type of problem but who don't understand the mechanism you would use to solve the problem without the rule or understand the limitations on the rule with respect to when it is or isn't applicable. I think the best rule is to only use higher-level methods to solve a problem when you already understand the "hard way" to the solution but simply need the answer - say as a step in an engineering application or in solving a more difficult set of equations. If this question is being asked from the context of a course which is teaching more fundamentals methods in mathematics, then while the L'Hopital's solution may be keen (and may even impress the teacher), it may be missing the more simplistic algebraic solution to the problem and teaching bad habits early.

Just a thought.
콩까지마
sigma_x
Profile Joined March 2008
Australia285 Posts
Last Edited: 2008-05-27 11:28:33
May 27 2008 11:07 GMT
#47
There are 3 primary ways of solving (limit) questions. 1) if the question is a polynomial over a polynomial, then the method with the least amount of effort is to simply divide numerator and denominator by the highest power. 2) L'Hopital's rule, only if it is in the right form (0/0 or infinity/infinity etc.) and you think it will get you somewhere - there are examples where L'Hopital will lead you into a circle, or something even more complicated. 3) Taylor (and other) Expansions.

The present question, however, is much easier than that because division by a common factor will yield the answer immediately.

Edit: There are other methods of course, the most well known being the squeeze theorem. Other more advanced methods require knowledge of set topology.
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