Answer
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - x + \dfrac{{{x^3}}}{6}}}{{{x^5}}} = \dfrac{1}{{120}}$
Work Step by Step
From Table 2 we have
$\sin x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
Hence, the limit becomes
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - x + \dfrac{{{x^3}}}{6}}}{{{x^5}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot}}{{{x^5}}} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{5!}} - \dfrac{{{x^2}}}{{7!}} + \cdot\cdot\cdot} \right)$
Thus, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - x + \dfrac{{{x^3}}}{6}}}{{{x^5}}} = \dfrac{1}{{5!}} = \dfrac{1}{{120}}$.
