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ksaimouli

  • one year ago

limts

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  1. ksaimouli
    • one year ago
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    http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__91.gif

  2. ksaimouli
    • one year ago
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    @tcarroll010

  3. tcarroll010
    • one year ago
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    You can use L'Hopital's rule whereby you take the limit of the derivative of the numerator over the limit of the derivative of the denominator. All with respect to "t".

  4. ksaimouli
    • one year ago
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    yes i did that but stuck

  5. tcarroll010
    • one year ago
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    Are you able to calculate the derivative of the numerator and the derivative of the denominator? You got stuck? Ok, I'll help a little further.

  6. ksaimouli
    • one year ago
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    so \[\sec(\frac{ \pi }{ 4 })^2-(\sec \frac{ \pi }{4 })^2\]

  7. ksaimouli
    • one year ago
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    ?

  8. ksaimouli
    • one year ago
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    do u know what is the answer it is 2

  9. ksaimouli
    • one year ago
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    but it is 2 i am sure

  10. tcarroll010
    • one year ago
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    The numerator goes to: sec^2 [(1/4)pi] as "t" goes to "0" So, yes, it goes to "2"

  11. tcarroll010
    • one year ago
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    Yes, definitely, "2" is the answer.

  12. ksaimouli
    • one year ago
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    how did u get that

  13. tcarroll010
    • one year ago
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    The derivative of the numerator is: sec^2 [(1/4)pi + t] and that goes to: sec^2 [(1/4)pi] as "t" goes to "0" The derivative of the denominator is just "1" So, you deal with just: sec^2 [(1/4)pi] and that = 2

  14. ksaimouli
    • one year ago
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    what did u do with other half sec^2 [(1/4)pi + t]- sec^2 [(1/4)pi )

  15. tcarroll010
    • one year ago
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    The other half is just a constant, so when you take the derivative of that, it disappears. That was where I made an initial mistake but I corrected that.

  16. ksaimouli
    • one year ago
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    chain rule does not apply for the first one

  17. tcarroll010
    • one year ago
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    Here's a graph of the function and you can see that the limit is "2".

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  18. tcarroll010
    • one year ago
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    The chain actually does apply to the first term, but it doesn't matter, because the derivative of (pi)/4 + t is just "1", so you are multiplying that derivative by "1", so it doesn't become apparent.

  19. ksaimouli
    • one year ago
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    alright thx

  20. tcarroll010
    • one year ago
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    Thx for the recognition btw.

  21. ksaimouli
    • one year ago
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    for long conversation

  22. tcarroll010
    • one year ago
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    And you're welcome. Sorry I went off on a tangent (no pun intended) at the beginning.

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