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

yairl · Feb 15, 2016

when i have lim x->0 (x-sinx)/(sinx)^3 why can't I replace it with 1/(sinx)^2-1/(sinx)^2?
thanks

Ssnow · Feb 15, 2016

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## ...

Ssnow · Feb 15, 2016

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

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

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