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ksaimouli

  • 2 years ago

limts

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

  2. ksaimouli
    • 2 years ago
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    @tcarroll010

  3. tcarroll010
    • 2 years 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
    • 2 years ago
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    yes i did that but stuck

  5. tcarroll010
    • 2 years 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
    • 2 years ago
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    so \[\sec(\frac{ \pi }{ 4 })^2-(\sec \frac{ \pi }{4 })^2\]

  7. ksaimouli
    • 2 years ago
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    ?

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

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

  10. tcarroll010
    • 2 years 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
    • 2 years ago
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    Yes, definitely, "2" is the answer.

  12. ksaimouli
    • 2 years ago
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    how did u get that

  13. tcarroll010
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    chain rule does not apply for the first one

  17. tcarroll010
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    alright thx

  20. tcarroll010
    • 2 years ago
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    Thx for the recognition btw.

  21. ksaimouli
    • 2 years ago
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    for long conversation

  22. tcarroll010
    • 2 years 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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