Question

Evaluate :
$\underset{x\to 0}{lim}\frac{x-\sin x\cos x}{x^3}.$

Answer 1

Solution -

${lim}_{x\to 0}\frac{x-sinxcosx}{x^3}$

$={lim}_{x\to 0}\frac{1-cosx}{x^2}$ because ${lim}_{x\to 0}\frac{sinx}{x}=1$

Apply L Hospital Rule twice

$={lim}_{x\to 0}\frac{sinx}{2x}$

$={lim}_{x\to 0}\frac{cosx}{2}$

$=\frac{1}{2}$