# When X tends to zero

1. Feb 15, 2016

### 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

2. Feb 15, 2016

### 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 $x\rightarrow 0$ ...

3. Feb 15, 2016

### Ssnow

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