Answer
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin \left( {{x^2}} \right)}}{{{x^4}}} - \dfrac{{\cos x}}{{{x^2}}}} \right) = \dfrac{1}{2}$
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 $x$
Substituting ${x^2}$ for $x$ gives
$\sin {x^2} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{{\left( {{x^2}} \right)}^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{4n + 2}}}}{{\left( {2n + 1} \right)!}} = {x^2} - \dfrac{{{x^6}}}{{3!}} + \dfrac{{{x^{10}}}}{{5!}} - \dfrac{{{x^{14}}}}{{7!}} + \cdot\cdot\cdot$
Also from Table 2:
$\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
Hence, the limit becomes
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin \left( {{x^2}} \right)}}{{{x^4}}} - \dfrac{{\cos x}}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{x^2} - \dfrac{{{x^6}}}{{3!}} + \dfrac{{{x^{10}}}}{{5!}} - \dfrac{{{x^{14}}}}{{7!}} + \cdot\cdot\cdot}}{{{x^4}}} - \dfrac{{1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot}}{{{x^2}}}} \right)$
$ = \mathop {\lim }\limits_{x \to 0} \left[ {\left( {\dfrac{1}{{{x^2}}} - \dfrac{{{x^2}}}{{3!}} + \dfrac{{{x^6}}}{{5!}} - \dfrac{{{x^{10}}}}{{7!}} + \cdot\cdot\cdot} \right) - \left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{2!}} + \dfrac{{{x^2}}}{{4!}} - \dfrac{{{x^4}}}{{6!}} + \cdot\cdot\cdot} \right)} \right]$
$ = \mathop {\lim }\limits_{x \to 0} \left[ {\left( {-\dfrac{{{x^2}}}{{3!}} + \dfrac{{{x^6}}}{{5!}} - \dfrac{{{x^{10}}}}{{7!}} + \cdot\cdot\cdot} \right) - \left( { - \dfrac{1}{{2!}} + \dfrac{{{x^2}}}{{4!}} - \dfrac{{{x^4}}}{{6!}} + \cdot\cdot\cdot} \right)} \right]$
$ = \dfrac{1}{{2!}}$
Thus, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin \left( {{x^2}} \right)}}{{{x^4}}} - \dfrac{{\cos x}}{{{x^2}}}} \right) = \dfrac{1}{2}$.
