L'Hospital's Rule

  • Thread starter tornzaer
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  • #1
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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.
 

Answers and Replies

  • #2
lanedance
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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....
 
  • #3
lanedance
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show your L'Hop working, i think it works fine, but has to be applied twice
 
  • #4
938
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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
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The first time applied gives me: (cosx - 1) / 3x^2

The second: (-sinx) / 6x

I definitely know this is wrong...
 
  • #6
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... and what would the third give you?
 
  • #7
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Third would be -cosx/6
 
  • #8
Gib Z
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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....

Pretty much what applying L'Hopital's rule does really.
 
  • #9
HallsofIvy
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Third would be -cosx/6
And the limit of that as x goes to 0 is ?
 
  • #10
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Try expanding sin(x) using the Maclaurin series.
You will get the answer directly.
 

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

That's not the answer according to calculators.
 
  • #12
525
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That's not the answer according to calculators.

According to Mathematica it is.
 
  • #13
lanedance
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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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