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sat_chen

  • one year ago

find this limit using the l hopital rule

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  1. sat_chen
    • one year ago
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    |dw:1361572558530:dw|

  2. sat_chen
    • one year ago
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    i know if you just plug in the value you get 0/0 so you use the rule

  3. sat_chen
    • one year ago
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    so i use the rule and get |dw:1361572627877:dw|

  4. sat_chen
    • one year ago
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    how do i get from here as you cant use the trigonometric substitution either

  5. walters
    • one year ago
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    |dw:1361572546361:dw|

  6. sat_chen
    • one year ago
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    the derivative of tan is sec^2 x lol

  7. sat_chen
    • one year ago
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    so thats wrong

  8. walters
    • one year ago
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    |dw:1361572818037:dw| then differentiate from here

  9. experimentX
    • one year ago
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    the limit goes to -infinity

  10. calmat01
    • one year ago
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    Change your 1 -sec^2x into -tan^2x and see if that helps you simplify your ratio.

  11. sat_chen
    • one year ago
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    thats not how it works youcan never change 1 - sec^2 into tan^2x

  12. calmat01
    • one year ago
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    you can change it into -tan^2x though, not tan^2x

  13. sat_chen
    • one year ago
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    ok im gonna try walters method

  14. calmat01
    • one year ago
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    But that only makes it worse. Nevermind. Hang on.

  15. sat_chen
    • one year ago
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    you dont understand tan^2 x + 1 = sec^2x try to change it to what you just did it never works

  16. calmat01
    • one year ago
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    Move the 1 over, and then multiply both sides by a -1.

  17. sat_chen
    • one year ago
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    i thought about your method walter but its x not 1 - tanx

  18. experimentX
    • one year ago
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    Without using L'hopital |dw:1361573289043:dw|

  19. sat_chen
    • one year ago
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    so the solution is 1/0 but thats not gonna work either

  20. experimentX
    • one year ago
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    or I could have misunderstood your question http://www.wolframalpha.com/input/?i=lim+x-%3E0+sin%28x%29%2F%28x-tan%28x%29%29

  21. calmat01
    • one year ago
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    If you use L'hopital's rule twice, you end up with the limit as x approaches negative infinity of cos^3x/(-2)

  22. sat_chen
    • one year ago
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    no that is the question experimentX

  23. calmat01
    • one year ago
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    oops, it's approaching zero, so the answer is -1/2

  24. walters
    • one year ago
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    |dw:1361573583223:dw|

  25. experimentX
    • one year ago
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    You can't use L'hopital rule twice, one you use L'hopital rule, ... the top is cos(x) ... which is 1 and the bottom part is still 0.

  26. experimentX
    • one year ago
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    *once

  27. calmat01
    • one year ago
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    Ah hell, then it's no longer of indeterminate form.

  28. walters
    • one year ago
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    |dw:1361573754561:dw|

  29. experimentX
    • one year ago
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    the best way see through limits is to expand using Taylor series. Try expanding it ... the answer is pretty obvious. Anyway gotta sleep ... best of luck.

  30. sat_chen
    • one year ago
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    i think i got it now ty a lot

  31. calmat01
    • one year ago
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    I need to go refresh my memory. Good luck to you.

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