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anonymous

  • 5 years ago

find the limit [(sin^2)x]/tanx as x approaches 0.

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  1. anonymous
    • 5 years ago
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    cos^2x=1

  2. anonymous
    • 5 years ago
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    whoops didnt see the tan x

  3. anonymous
    • 5 years ago
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    rewrite as sin^2(x)/[sinx/cosx] = sin(x)cos(x) The limit of sin(x)cos(x) can be thought of as the product of the limit of sin(x) and cos(x) separately. sin(x) approaches 0 as x apparoaches 0, and cos(x) approaches 1 as x apparoches 0, so the limit of sin(x)cos(x) = 0(1) = 0

  4. anonymous
    • 5 years ago
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    sin^2x/tanx = cos2x/sec^2x= cos^2x*cos^2x= 1 using L' Hospital's rule. Can you use that rule pyeh?

  5. anonymous
    • 5 years ago
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    Yes you can, but it's easier to rewrite.

  6. anonymous
    • 5 years ago
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    Also, derivative of sin^2(x) = 2sin(x)cos(x)

  7. anonymous
    • 5 years ago
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    why did i get 1 using that rule, and you got 0?

  8. anonymous
    • 5 years ago
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    see the above post; I think you took the derivative of the numerator incorrectly.

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