# L'Hospital's Rule

1. Nov 12, 2009

### tornzaer

1. The problem statement, all variables and given/known data

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

2. Relevant equations

L'Hospital's Rule

3. 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.

2. Nov 12, 2009

### 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....

3. Nov 12, 2009

### lanedance

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

4. Nov 12, 2009

### 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.

5. Nov 12, 2009

### tornzaer

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

The second: (-sinx) / 6x

I definitely know this is wrong...

6. Nov 12, 2009

### clamtrox

... and what would the third give you?

7. Nov 12, 2009

### tornzaer

Third would be -cosx/6

8. Nov 12, 2009

### Gib Z

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

9. Nov 12, 2009

### HallsofIvy

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

10. Nov 12, 2009

### blitz.km

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

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11. Nov 12, 2009

### tornzaer

That's not the answer according to calculators.

12. Nov 12, 2009

### xepma

According to Mathematica it is.

13. Nov 12, 2009

### lanedance

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...