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

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (sin(x) - x) / x^3 as x approaches 0, which presents an indeterminate form of 0/0. Participants are exploring the application of L'Hospital's Rule and considering alternative approaches such as Taylor series expansion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of L'Hospital's Rule and the necessity of rewriting the expression before applying it. There are questions about the correctness of the results obtained after multiple applications of the rule. Some suggest using Taylor series for a deeper understanding of the limit.

Discussion Status

The discussion is active, with participants sharing their attempts at applying L'Hospital's Rule and questioning the results. There is no explicit consensus on the correct approach, but several lines of reasoning are being explored, including the potential for misunderstanding in the application of the rule and the utility of Taylor series.

Contextual Notes

Some participants note that calculators provide different answers, leading to confusion about the correct limit. There are mentions of needing to show work and clarify methods to facilitate better assistance.

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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.
 
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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...
 
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
 
lanedance said:
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.
 
tornzaer said:
Third would be -cosx/6
And the limit of that as x goes to 0 is ?
 
  • #10
Try expanding sin(x) using the Maclaurin series.
You will get the answer directly.
 

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  • #11
HallsofIvy said:
And the limit of that as x goes to 0 is ?

That's not the answer according to calculators.
 
  • #12
tornzaer said:
That's not the answer according to calculators.

According to Mathematica it is.
 
  • #13
tornzaer said:
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...
 

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