Math Help! (Limits)

[EternalWrath]

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.

[bluguerrilla]

Re: Math Help! (Limits)

11/2, feh.

L'Hospital wins again.

Edit: Magus ftw.

[HeZe]

Re: Math Help! (Limits)

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

[MRichards]

Re: Math Help! (Limits)

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

[bluguerrilla]

Re: Math Help! (Limits)

maths

You..... badger!

[MrRubix]

Re: Math Help! (Limits)

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.

[EternalWrath]

Re: Math Help! (Limits)

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.

[rpozas19]

Re: Math Help! (Limits)

i think your right man i hate long divison problems