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

Question
Limits and continuity
The limit as x approaches zero of $$\displaystyle\frac{{{\sin{{x}}}-{x}{\cos{{x}}}}}{{x}^{{3}}}$$

2021-03-05
Let $$\displaystyle{f{{\left({x}\right)}}}=\frac{{g{{\left({x}\right)}}}}{{h}}{\left({x}\right)},{w}{h}{e}{r}{e}{g{{\left({x}\right)}}}={\sin{{x}}}-{x}.{\cos{{x}}},{\quad\text{and}\quad}{h}{\left({x}\right)}={x}^{{3}}.$$
The limit is $$\displaystyle{x}\to{0}.$$
Since g(0) and h(0) both = 0,then use l'Hopital's rule, $$\displaystyle\lim{f{{\left({x}\right)}}}{\left[{x}\to{0}\right]}=\Lim{\left[{x}\to{0}\right]}{g}'\frac{{{x}}}{{h}}'{\left({x}\right)}$$
$$\displaystyle{g}'{\left({x}\right)}={\cos{{x}}}-{\cos{{x}}}+{x}.{\sin{{x}}}={x}.{\sin{{x}}}$$
$$\displaystyle{h}'{\left({x}\right)}={3}{x}^{{2}}$$
Since g'(0) and h'(0) both = 0,then again use l'Hopital's rule, i.e. $$\displaystyle\lim{f{{\left({x}\right)}}}{\left[{x}\to{0}\right]}=\Lim{\left[{x}\to{0}\right]}{g}{''}\frac{{{x}}}{{h}}{''}{\left({x}\right)}$$
$$\displaystyle{g}{''}{\left({x}\right)}={\sin{{x}}}+{x}.{\cos{{x}}}$$
$$\displaystyle{h}{''}{\left({x}\right)}={6}{x}$$
Since g''(0) and h''(0) both = 0,then again use l'Hopital's rule, i.e. $$\displaystyle\lim{f{{\left({x}\right)}}}{\left[{x}\to{0}\right]}=\Lim{\left[{x}\to{0}\right]}{g}{'''}\frac{{{x}}}{{h}}{'''}{\left({x}\right)}$$
$$\displaystyle{g}{'''}{\left({x}\right)}={\cos{{x}}}+{\cos{{x}}}-{x}.{\sin{{x}}}={2}{\cos{{x}}}-{x}.{\sin{{x}}}$$
$$\displaystyle{h}{'''}{\left({x}\right)}={6}$$
Now, g''(0) = 2, and h'''(0) = 6.
So, $$\displaystyle\lim{f{{\left({x}\right)}}}{\left[{x}\to{0}\right]}=\Lim{\left[{x}\to{0}\right]}{g}{'''}\frac{{{x}}}{{h}}{'''}{\left({x}\right)}=\Lim{\left[{x}\to{0}\right]}\frac{{2}}{{6}}=\frac{{1}}{{3}}$$

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