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

The limit as x approaches zero of $\frac{\mathrm{sin}x-x\mathrm{cos}x}{{x}^{3}}$
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rogreenhoxa8

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