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xilinix

  • 2 years ago

lim -> 0 sin(2x) /(x*cos(x) ) help

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  1. myko
    • 2 years ago
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    sin (2x) = 2sin(x)cos(x) so sin(2x) /(x*cos(x) =2sin(x)cos(x)/x*cos(x)=2sinx/x which as x->0 goes to 1

  2. xilinix
    • 2 years ago
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    no, answer is 2

  3. myko
    • 2 years ago
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    ups sry, forgot about 2 multuplying: lim 2sinx/x=2*1=2 x->0

  4. xilinix
    • 2 years ago
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    don't undestand you

  5. myko
    • 2 years ago
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    sin (2x) can be written like: sin (2x) = 2sin(x)cos(x) substituting this expretion in the original one: sin(2x) /(x*cos(x) =2sin(x)cos(x)/x*cos(x)=2sinx/x taking limits, and noticing that lim as x->0 of sinx/x=1: lim 2sinx/x=2*1=2 x->0

  6. xilinix
    • 2 years ago
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    |dw:1352667129469:dw|

  7. myko
    • 2 years ago
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    no

  8. xilinix
    • 2 years ago
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    why not? l'Hôpital rule

  9. ccappiello
    • 2 years ago
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    yes, using l'hospital's rule works

  10. Varuns
    • 2 years ago
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    Don't break your head ! You know that sin(x)=x as x tends to zero. So sin(2x) = 2x. In the denominator, keep x as it is and cos(x)=1. Hence the answer

  11. xilinix
    • 2 years ago
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    thanks, but the image is ok?

  12. Varuns
    • 2 years ago
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    |dw:1352669528245:dw|

  13. lopus
    • 2 years ago
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    claro que es posible es solamente derivar arriba y abajo como lo muestras en la imagen y ya

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