Exercise:
Show that $\underset{x\to
0}{\mathop{\lim }}\,\frac{x-\sin x}{{{x}^{3}}}=\frac{1}{6}$
Solution:
we
know that $\sin 2x=2\sin x\cos x\Rightarrow \sin x=2\sin \left( \frac{x}{2}
\right)\cos \left( \frac{x}{2} \right)$
So $\frac{x-\sin
x}{{{x}^{3}}}=\frac{x-2\sin \left( \frac{x}{2} \right)\cos x-2\sin \left(
\frac{x}{2} \right)+2\sin \left( \frac{x}{2} \right)}{{{x}^{3}}}$
$=\frac{x-2\sin
\left( \frac{x}{2} \right)}{{{x}^{3}}}+\frac{2\sin \left( \frac{x}{2}
\right)\left( 1-\cos \left( \frac{x}{2} \right) \right)}{{{x}^{3}}}$
But $\underset{x\to
0}{\mathop{\lim }}\,\frac{x-2\sin \left( \frac{x}{2}
\right)}{{{x}^{3}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{x-2\sin
\left( \frac{x}{2} \right)}{8}}{\frac{{{x}^{3}}}{8}}$ $=\underset{x\to
0}{\mathop{\lim }}\,\frac{\frac{1}{4}\left( \frac{x}{2}-\sin \left( \frac{x}{2}
\right) \right)}{{{\left( \frac{x}{2} \right)}^{3}}}$ $=\frac{1}{4}\underset{x\to
0}{\mathop{\lim }}\,\frac{\frac{x}{2}-\sin \left( \frac{x}{2} \right)}{{{\left(
\frac{x}{2} \right)}^{3}}}$
Let $u=\frac{x}{2}$
as $x\to 0$ ,$u\to 0$ $\Rightarrow \underset{x\to 0}{\mathop{\lim
}}\,\frac{\frac{x}{2}-\sin \left( \frac{x}{2} \right)}{{{\left( \frac{x}{2}
\right)}^{3}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{u-\sin u}{{{u}^{3}}}$$=\underset{x\to
0}{\mathop{\lim }}\,\frac{x-\sin x}{{{x}^{3}}}$
also We have
$\frac{2\sin
\left( \frac{x}{2} \right)\left( 1-\cos \left( \frac{x}{2} \right)
\right)}{{{x}^{3}}}=\frac{8\cos \left( \frac{x}{4} \right){{\sin }^{3}}\left(
\frac{x}{4} \right)}{{{x}^{3}}}$
$=8\cos \left( \frac{x}{4} \right)\times \frac{{{\sin
}^{3}}\left( \frac{x}{4} \right)}{{{x}^{3}}}$ But $\underset{x\to
0}{\mathop{\lim }}\,8\cos \left( \frac{x}{4} \right)=8$
And
$\underset{x\to 0}{\mathop{\lim
}}\,\frac{{{\sin }^{3}}\left( \frac{x}{4} \right)}{{{x}^{3}}}=\underset{x\to
0}{\mathop{\lim }}\,\frac{\frac{1}{64}{{\sin }^{3}}\left( \frac{x}{4}
\right)}{{{\left( \frac{x}{4} \right)}^{3}}}=\frac{1}{64}$
Thus $\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin
x}{{{x}^{3}}}=\frac{1}{4}\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin
x}{{{x}^{3}}} \right)+\frac{8}{64}$ $\Leftrightarrow \frac{3}{4}\underset{x\to
0}{\mathop{\lim }}\,\frac{x-\sin x}{{{x}^{3}}}=\frac{8}{64}$
$\Rightarrow
\underset{x\to 0}{\mathop{\lim }}\,\frac{x-\sin
x}{{{x}^{3}}}=\frac{8}{64}\times \frac{4}{3}=\frac{1}{6}$