Answer
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^{ - 1}}x - x\cos x - \dfrac{1}{6}{x^3}}}{{{x^5}}} = \dfrac{{19}}{{120}}$
Work Step by Step
From Table 2 we have
${\tan ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{2n + 1}} = x - \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} - \dfrac{{{x^7}}}{7} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \le 1$
$\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
So,
$x\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n} \right)!}} = x - \dfrac{{{x^3}}}{{2!}} + \dfrac{{{x^5}}}{{4!}} - \dfrac{{{x^7}}}{{6!}} + \cdot\cdot\cdot$
Hence, the limit becomes
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^{ - 1}}x - x\cos x - \dfrac{1}{6}{x^3}}}{{{x^5}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {x - \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} - \dfrac{{{x^7}}}{7} + \cdot\cdot\cdot} \right) - \left( {x - \dfrac{{{x^3}}}{{2!}} + \dfrac{{{x^5}}}{{4!}} - \dfrac{{{x^7}}}{{6!}} + \cdot\cdot\cdot} \right) - \dfrac{1}{6}{x^3}}}{{{x^5}}}$
Collecting terms with same orders, we get
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( { - \dfrac{{{x^3}}}{3} + \dfrac{{{x^3}}}{2} - \dfrac{1}{6}{x^3}} \right) + \left( {\dfrac{{{x^5}}}{5} - \dfrac{{{x^5}}}{{4!}}} \right) - \left[ {\left( { - \dfrac{{{x^7}}}{{6!}} + \dfrac{{{x^7}}}{7}} \right) + \cdot\cdot\cdot} \right]}}{{{x^5}}} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{5} - \dfrac{1}{{4!}}} \right) - \left[ {\left( { - \dfrac{{{x^2}}}{{6!}} + \dfrac{{{x^2}}}{7}} \right) + \cdot\cdot\cdot} \right]$
Thus, $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^{ - 1}}x - x\cos x - \dfrac{1}{6}{x^3}}}{{{x^5}}} = \dfrac{1}{5} - \dfrac{1}{{24}} = \dfrac{{19}}{{120}}$.
