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Find Limit

  1. Jan 4, 2008 #1
    Find [tex]\mathop {\lim }\limits_{x \to 0} \frac{{\tan (nx) - n\tan (x)}}
    {{n\sin (x) - \sin (nx)}} [/tex]
     
  2. jcsd
  3. Jan 4, 2008 #2

    CompuChip

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    This looks like homework.
    What did you try already?
    Taylor expansion? L'Hôpitals rule?
     
  4. Jan 4, 2008 #3
    I'm taking a guess here, but couldn't tan(nx) be extracted to sin(nx) / cos(nx) ?

    Or is that complicating things, or just wrong?
     
  5. Jan 4, 2008 #4

    CompuChip

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    That's right and possible though I doubt it will simplify things (but unless you know the Taylor expansion / small argument approximation / derivative and what more you may need to calculate the limit for the tangent, you can use tan = sin/cos to find them). But I think it is time for the TS to show some work.
     
    Last edited: Jan 4, 2008
  6. Jan 4, 2008 #5

    Gib Z

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    Small angle approximation, Taylor expansion and L'Hopital's rule is all exactly the same :( Well, at least here it is.
     
  7. Jan 4, 2008 #6

    dynamicsolo

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    Another approach, after rewriting the tangent terms as sine/cosine terms, would be to divide top and bottom by x and, with multiplications above and below by n in appropriate places, exploit the heck out of

    lim u->0 (sin u)/u = 1 .

    [EDIT: Ah-haha! Not so simple as this one looks. This approach does nothing to fix the problem with the denominator. And l'Hopital doesn't fare much better, as you'll get 0 - 0 in the denominator endlessly. I'm checking to see if there's a "nice" way to eliminate the difference in the denominator...]
     
    Last edited: Jan 5, 2008
  8. Jan 5, 2008 #7

    Gib Z

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    Which, again, is just another form of applying small angle approximation/l'hopital/taylor series =D
     
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