|
|
#1 |
|
FFR Player
|
All right, so apparently my math teacher is in cahoots with these other math teachers from other schools, and one of them had difficulty solving a particular limit (yes, the teacher had trouble). My teacher said she didn't have any trouble, but she thought that if it was difficult enough to give a teacher a headache, then it is an excellent problem to give the class to solve.
![]() Here's the problem: "Find the limit as x approaches 1 of the function: [(3x^2-4)sqrt(x) + 1] / (x-1)" I'm going to try to solve it on my own, but if for some reason I can't get it I will check on here later for the answer.
__________________
![]() Your defenestration is imminent Like video games, but are tired of talking with the same noobs online? Check out The eLounge. If you're not 100% satisfied, then... um... just leave. |
|
|
|
|
|
#2 |
|
This is a custom title.
|
11/2, feh.
L'Hospital wins again. Edit: Magus ftw.
__________________
|
|
|
|
|
|
#3 | |
|
FFR Veteran
|
Lol.. Our advanced math teacher taught this today but this was just too hard. I tried twice but I got just (9x^5-16x+1)/[(x-1)(3x^(5/2)-4x^(1/2)+1)].. lol not going further. Good luck
EDIT: Dam bluguerrilla GJ
__________________
![]() Quote:
|
|
|
|
|
|
|
#4 |
|
FFR Player
Join Date: May 2006
Posts: 91
|
lim ((3x^2 - 4).sqrt(x) + 1) / (x - 1)
x->1 = lim (3x^2.5 - 4x^0.5 + 1) / (x - 1) [1] 1. With L'Hospital's rule: [lim f(x)/g(x) = lim f'(x)/g'(x)] [1] = lim (7.5x^1.5 - 2x^-0.5) / 1 = (7.5 - 2) / 1 = 5.5 2. Without L'Hospital: Long division of (x-1) into left part of numerator [1] = lim ((x-1)(3x^1.5 + 3x^0.5) - x^0.5 + 1) / (x - 1) Simplify left part and factorise remaining denominator: = lim 3x^1.5 + 3x^0.5 - (x^0.5 - 1) / ((x^0.5 + 1)(x^0.5 - 1)) = lim 3x^1.5 + 3x^0.5 - 1 / (x^0.5 + 1) = 3 + 3 - 0.5 = 5.5 Edit: Excuse the .5's rather than /2's - I'm a programmer ![]() Last edited by MRichards; 10-8-2007 at 11:14 AM.. |
|
|
|
|
|
#5 |
|
This is a custom title.
|
__________________
|
|
|
|
|
|
#6 |
|
FFR Player
Join Date: Dec 1969
Location: New York City, New York
Posts: 8,023
|
You've gotta be kidding me, that limit is easy -- the teacher had trouble!? L'Hospital's Rule is obviously the quick fix to such a limit.
Kudos to MRichards for the long division, that is pretty slick. |
|
|
|
|
|
#7 |
|
FFR Player
|
First of all, I haven't learned L'Hospital's Rule yet (wait, I have now!
) but even if I did know it I don't think the teacher would allow us to use it on this problem since she hasn't taught us it. That still doesn't excuse the other math teacher... oh well, maybe he was trying to solve it without L'Hospital's Rule. ![]() I think the method MRichards proposed will do nicely. Up till now I worked my way to the equation [(x+1)(3x^2-3x-1)] / [(x-1)(x^.5-1)] and I don't think I can go anywhere useful from there. Long division works much better. Oh, and yes, Magus ftw. ![]()
__________________
![]() Your defenestration is imminent Like video games, but are tired of talking with the same noobs online? Check out The eLounge. If you're not 100% satisfied, then... um... just leave. |
|
|
|
|
|
#8 |
|
Banned
Join Date: Oct 2007
Posts: 2
|
i think your right man i hate long divison problems
|
|
|
|
![]() |
| Currently Active Users Viewing This Thread: 1 (0 members and 1 guests) | |
|
|