How do you solve (1 - (cosx)^3)/x^2 as x -> 0? I tried to use l'Hospital rule but it didn't help.
You can calculate this limit without L'Hospital's rule:
$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}x}}
{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \left( {1 - {{\sin }^2}x} \right)\cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x + {{\sin }^2}x\cos x}}{{{x^2}}} =$
$\displaystyle = 2\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}
{2x^2} + \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}
{{{x^2}}}\cos x = 2\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}} + {\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}} \right)^2}\mathop {\lim }\limits_{x \to 0} \cos x =$
$\displaystyle = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{\frac{{{x^2}}}{4}}} + 1 = \frac{1}{2}{\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)^2} + 1 = \frac{1}{2} + 1 = \frac{3}{2}.$