Lim x->0: Why Can't I Replace (x-sinx)/(sinx)^3 with 1/(sinx)^2-1/(sinx)^2?

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SUMMARY

The limit as x approaches 0 of (x - sin x) / (sin x)^3 cannot be simplified to 1/(sin x)^2 - 1/(sin x)^2 due to the incorrect application of the limit properties. The discussion emphasizes the necessity of using a more accurate approximation than sin x ~ x when evaluating limits. Specifically, the use of L'Hôpital's rule or the Maclaurin series expansion is recommended for precise calculations in this context.

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when i have lim x->0 (x-sinx)/(sinx)^3 why can't I replace it with 1/(sinx)^2-1/(sinx)^2?
thanks
 
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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## ...
 
You know the de L'Hopital rule or Mc Laurin expansion?
 

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