Solve L'Hospital's Rule Homework with 0/0

[tornzaer]

Homework Statement

lim -> 0 (sin(x)-x) / x^3

Homework Equations

L'Hospital's Rule

The Attempt at a Solution

I could just use L'Hospital's rule since it's 0/0. However, the answer is wrong when I do it that way. What am I missing? It states on the question that it must be rewritten before the ruel can be applied and that it has to be applied more than once.

Please help.

 

[lanedance]

will have a closer look, but one way to gain better understading about the problem would be to expand the sin in a taylor series about 0...

 

[lanedance]

show your L'Hop working, i think it works fine, but has to be applied twice

 

[clamtrox]

I don't see why you'd need to rewrite anything. You just have to argue that

lim x->0 f/g = lim x->0 f'/g' = lim x->0 f''/g'' = ... is true for the nth derivate as long as the limit of n-1th derivate is either zero or infinite for both f and g.

 

[tornzaer]

The first time applied gives me: (cosx - 1) / 3x^2

The second: (-sinx) / 6x

I definitely know this is wrong...

 

[clamtrox]

... and what would the third give you?

 

[tornzaer]

Third would be -cosx/6

 

[Gib Z]

will have a closer look, but one way to gain better understading about the problem would be to expand the sin in a taylor series about 0...

Pretty much what applying L'Hopital's rule does really.

 

[HallsofIvy]

Third would be -cosx/6

And the limit of that as x goes to 0 is ?

 

[blitz.km]

Try expanding sin(x) using the Maclaurin series.
You will get the answer directly.

 

[tornzaer]

And the limit of that as x goes to 0 is ?

That's not the answer according to calculators.

 

[xepma]

That's not the answer according to calculators.

According to Mathematica it is.

 

[lanedance]

That's not the answer according to calculators.

statements like that, with very little information, are hard to help with...

show your method & results and explain why you think there is a disconnect, and we can comment/help out...