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anonymous

  • 5 years ago

limit of (2x(csc4x)^2) as x approaches 0

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  1. anonymous
    • 5 years ago
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    2x is zero. Brush up on a lil trig what is csc 0?

  2. anonymous
    • 5 years ago
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    1/sin(0)

  3. anonymous
    • 5 years ago
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    If 2x is zero wouldn't it all be zero?

  4. anonymous
    • 5 years ago
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    In a finite world, yes, but you are currently residing in a potentially infinite world: possibility of 0 times infinity=indeterminate.

  5. anonymous
    • 5 years ago
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    I think the answer is undefined not indeterminate, there is a difference. If you try to solve using l'hospital's rule, you end up getting an answer of 2/0 when plugging in 0. 2/0 is undefined not indeterminate. For example, one indeterminate form is 0/0.

  6. anonymous
    • 5 years ago
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    It may not get to that point. According to her finding 1/sin (0) is 1. Indeterminate was just mentioned as a potential, the need to investigate csc 0.

  7. anonymous
    • 5 years ago
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    when using l'hospital's rule don't we take the derivative of each or is that wrong?

  8. anonymous
    • 5 years ago
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    I don't know that you need. It seems the lim is 0 times 1; zero

  9. anonymous
    • 5 years ago
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    Oh no, sin is 0, yeah lhopital

  10. anonymous
    • 5 years ago
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    I think it is chain rule. I think it is only when one on top, one on bottom, you do it the other way.

  11. anonymous
    • 5 years ago
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    Yeah i'm thinking this specific problem is chain rule

  12. anonymous
    • 5 years ago
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    You can use l'hospital when you come across indeterminate forms. For example, in this problem plugging in 0 gives us 2(0)/[sin(4(0))]^2 which is equal to 0/0, which is an indeterminate form. Consequently, we can use l'hospital to investigate whether or not the limit is solvable or undefined. We do this by taking the derivative of the top and the derivative of the bottom, which yields: 2/8sin(4x)cos(4x). If we now plug in o we get 2/8sin(4(0))cos(4(0)), which equals 2/0. 2/0 is undefinded and we cannot perform l'hospitals rule anymore; the answer is just undefined..i think anyway.

  13. anonymous
    • 5 years ago
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    Okay, that makes sense. Thank you!

  14. anonymous
    • 5 years ago
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    Yep, no problem!

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