Answer
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x - 1 + \dfrac{{{x^2}}}{2}}}{{{x^4}}} = \dfrac{1}{{24}}$
Work Step by Step
From Table 2 we have
$\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
Hence, the limit becomes
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x - 1 + \dfrac{{{x^2}}}{2}}}{{{x^4}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot}}{{{x^4}}} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{4!}} - \dfrac{{{x^2}}}{{6!}} + \cdot\cdot\cdot} \right)$
Thus, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x - 1 + \dfrac{{{x^2}}}{2}}}{{{x^4}}} = \dfrac{1}{{4!}} = \dfrac{1}{{24}}$.
