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

[yairl]

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]

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]

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