I admit, my calculus is a bit fuzzy after not having it in over 8 years. But, since my girlfriend is taking a Calculus class I've found that I know a lot more about math and calculus then I thought I did. So far I've been able to answer every question she's had, cept for this one. When all else fails, turn to the Home Theater Forum collective for help.... so here goes. The problem ask for the values of a and b so that: Limit [(ax-b)^(1/2) - 3] / x = 2 (x->0) If I can get her the answer to this problem... she's promised me conjugal relations. Please help a brother out!

Looks like we have a notation difference between US & UK. So I'm assuming you're taking the differential of the term in square brackets and x is greater than 0? Multiply both sides by x. Integrate the rhs. Square both sides. Reorder. I get x^4 + 6x^2 -ax + ( 9 + b ) = 0 So there's Avi's b=-9

Avi's method should get you a correct answer. I'm pretty sure it is the only one, not one of an infinite set like he suggests. I get a=12 and b=-9 If anyone can post different values that work, I would be surprised

I also get a=12 and b= -9 The work: (assume lim = lim as x approaches 0) using l'hospital's rule : lim ½a(ax-b)^(-½) = 2 lim a(ax-b)^(-½) = 4 a(-b)^(-½) = 4 a/(-b^½) = 4 12/3 = 4 yay! now have some fun with your amiga

I got a = 12 and b = -9 ---- (((ax - b)^1/2) - 3)/x = 2 (ax - b)^1/2 - 3 = 2x (ax - b)^1/2 = 2x + 3 (ax - b) = (2x + 3)^2 (ax - b) = 4x^2 + 12x + 9 lim x -> 0 b = -9 Substituting b = -9 and resolving: ax + 9 = 4x^2 + 12x + 9 ax = 4x^2 + 12x a = 4x + 12 lim x ->0 a = 12