Question

The limit as x approaches zero of \(\displaystyle\frac{{{\sin{{x}}}-{x}{\cos{{x}}}}}{{x}^{{3}}}\)

Answer 1

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, \(\displaystyle\lim{f{{\left({x}\right)}}}{\left[{x}\to{0}\right]}=\Lim{\left[{x}\to{0}\right]}{g}{'''}\frac{{{x}}}{{h}}{'''}{\left({x}\right)}=\Lim{\left[{x}\to{0}\right]}\frac{{2}}{{6}}=\frac{{1}}{{3}}\)