# L'Hospital's Rule

## 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.

## Answers and Replies

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

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.

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

The second: (-sinx) / 6x

I definitely know this is wrong...

... and what would the third give you?

Third would be -cosx/6

Gib Z
Homework Helper
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
Science Advisor
Homework Helper
Third would be -cosx/6
And the limit of that as x goes to 0 is ?

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

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

That's not the answer according to calculators.

That's not the answer according to calculators.

According to Mathematica it is.

lanedance
Homework Helper
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...