The limit as x approaches zero of (sinx - xcosx)/x^3

Let ${\displaystyle f\left(x\right)=\frac{g\left(x\right)}{h}\left(x\right),whereg\left(x\right)=\sin x-x.\cos x,\text{and}h\left(x\right)=x^3.}$
The limit is ${\displaystyle x\to 0.}$
Since g(0) and h(0) both $=0$,then use l'Hopital's rule, $limf(x)[x->0]=lim[x->0]g^{″}(x)/h^{″}(x)$
${\displaystyle g^{\prime }\left(x\right)=\cos x-\cos x+x.\sin x=x.\sin x}$
${\displaystyle h^{\prime }\left(x\right)=3x^2}$
Since g'(0) and h'(0) both = 0,then again use l'Hopital's rule, i.e. $limf(x)[x->0]=lim[x->0]g^{″}(x)/h^{″}(x)$
${\displaystyle g{}^{″}\left(x\right)=\sin x+x.\cos x}$
${\displaystyle h{}^{″}\left(x\right)=6x}$
Since g''(0) and h''(0) both = 0,then again use l'Hopital's rule, i.e. $limf(x)[x->0]=lim[x->0]g^{‴}(x)/h^{‴}(x)$
${\displaystyle g{}^{‴}\left(x\right)=\cos x+\cos x-x.\sin x=2\cos x-x.\sin x}$
${\displaystyle h{}^{‴}\left(x\right)=6}$
Now, $g^{″}(0)=2$, and $h^{‴}(0)=6$.
So, $limf(x)[x->0]=lim[x->0]g^{‴}(x)/h^{‴}(x)=lim[x->0]2/6=1/3$