1. ## Limit

How do you solve (1 - (cosx)^3)/x^2 as x -> 0? I tried to use l'Hospital rule but it didn't help.

use it two times and you can get the answer

3. you only need L'Hopitals Rule once:

(1 - (cosx)^3)/x^2

differentiate:

3cos^2(x) {sin(x)/2x}

and

lim 3cos^2(x) {sin(x)/2x} = 3/2 as lim sin(x)/x = 1

4. Originally Posted by antero
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:

$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}x}} {{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \left( {1 - {{\sin }^2}x} \right)\cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x + {{\sin }^2}x\cos x}}{{{x^2}}} =$

$\displaystyle = 2\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}} {2x^2} + \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}} {{{x^2}}}\cos x = 2\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}} + {\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}} \right)^2}\mathop {\lim }\limits_{x \to 0} \cos x =$

$\displaystyle = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{\frac{{{x^2}}}{4}}} + 1 = \frac{1}{2}{\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)^2} + 1 = \frac{1}{2} + 1 = \frac{3}{2}.$