Question

$\underset{x\to 0}{lim}\frac{x-\sin x}{x^3}$

Answer 1

$\underset{x\to 0}{lim}\frac{x-\sin x}{x^3}$

Let $f\left(x\right)=x-\sin \left(x\right)$

$g\left(x\right)=x^3$

By $L^{\prime }$Hospital Rule

$f^{\prime }\left(x\right)=1-\cos x$

$g^{\prime }\left(x\right)=3x^2$

$\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=\infty$

again differemtiate

$f^{\prime \prime }\left(x\right)=+\sin x$

$g"\left(x\right)=6x$

$\underset{x\to 0}{lim}\frac{\sin x}{6x}=\frac{1}{6}\underset{x\to 0}{lim}\frac{\sin x}{x}$

$=\frac{1}{6}$