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tamiashi

  • one year ago

Find the limit lim {x->0} (1-cos6x)/(x sin5x) How to solve this?

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  1. completeidiot
    • one year ago
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    are you allowed to use lhopital?

  2. TuringTest
    • one year ago
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    everybody jumps for l'hospital... that should really be a last resort

  3. TuringTest
    • one year ago
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    is this\[\lim_{x\to0}{1-\cos^6x\over x\sin^5x}\]?

  4. abb0t
    • one year ago
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    I think it would make this easier to use l'hopitals rule for this on an exam though, just to save time.

  5. TuringTest
    • one year ago
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    Often you are not allowed to use l'hospital if the problem is\[\lim_{x\to0}{1-\cos(6x)\over x\sin(5x)}\]it can be solved fairly easily without l'hospital. so far, however, @tamiashi has not replied to me, so I don't want to go ahead with either until I get a response.

  6. tamiashi
    • one year ago
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    Sorry for being late in response, I'm not allowed to use L'hospital Can you show me how to solve it without L'hospital?

  7. TuringTest
    • one year ago
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    is it the first way I wrote it or the second? are 5 and 6 exponents, or coefficients of the argument?

  8. tamiashi
    • one year ago
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    it's the second way

  9. TuringTest
    • one year ago
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    first multiply by\[\frac{1+\cos(6x)}{1+\cos(6x)}\]

  10. TuringTest
    • one year ago
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    you then have\[{1-\cos^2(6x)\over x\sin(5x)(1+\cos(6x))}=\frac{1-\cos^2(6x)}{x}\cdot\frac1{\sin(5x)}\cdot\frac1{1+\cos(6x)}\]now multiply by \(\frac xx\) to get\[{1-\cos^2(6x)\over x\sin(5x)(1+\cos(6x))}=\frac{1-\cos^2(6x)}{x^2}\cdot\frac x{\sin(5x)}\cdot\frac1{1+\cos(6x)}\] manipluate this through trig identities and algebra sp that you can utilize\[\lim_{x\to0}\frac{\sin x}x=1\]to get your answer.

  11. tamiashi
    • one year ago
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    okay got it now, thank you so much ^^

  12. TuringTest
    • one year ago
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    welcome :D

  13. RadEn
    • one year ago
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    alternative : use the identity : sin^2 (3x) = (1-cos(6x))/2 ---> 1 - cos(6x) = 2 sin^2 (3x) so, it can be : |dw:1361606909900:dw|

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