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When X tends to zero

  1. Feb 15, 2016 #1
    when i have lim x->0 (x-sinx)/(sinx)^3 why can't I replace it with 1/(sinx)^2-1/(sinx)^2?
    thanks
     
  2. jcsd
  3. Feb 15, 2016 #2

    Ssnow

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    because ##\frac{x-\sin{x}}{(\sin{x})^3}\not=\frac{1}{\sin^{2}{x}}-\frac{1}{\sin^{2}{x}}##. I understand that you want to use the limit ##\lim_{x\rightarrow 0}\frac{\sin{x}}{x}=1##, but in this case you need a better approximation than ##\sin{x}\sim x## as ##x\rightarrow 0## ...
     
  4. Feb 15, 2016 #3

    Ssnow

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    You know the de L'Hopital rule or Mc Laurin expansion?
     
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