Answer
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}{{\rm{e}}^x}}}{{\cos x - 1}} = - 2$
Work Step by Step
From Table 2 of Section 11.7, we have
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{6} + \dfrac{{{x^4}}}{{24}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
$\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
Thus, we get
${x^2}{{\rm{e}}^x} = {x^2} + {x^3} + \dfrac{{{x^4}}}{2} + \dfrac{{{x^5}}}{6} + \dfrac{{{x^6}}}{{24}} + \cdot\cdot\cdot$
$\cos x - 1 = - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$
Hence, the limit becomes
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}{{\rm{e}}^x}}}{{\cos x - 1}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} + {x^3} + \dfrac{{{x^4}}}{2} + \dfrac{{{x^5}}}{6} + \dfrac{{{x^6}}}{{24}} + \cdot\cdot\cdot}}{{ - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot}}$
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{6} + \dfrac{{{x^4}}}{{24}} + \cdot\cdot\cdot}}{{ - \dfrac{1}{{2!}} + \dfrac{{{x^2}}}{{4!}} - \dfrac{{{x^4}}}{{6!}} + \cdot\cdot\cdot}} = - 2$
So, $\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}{{\rm{e}}^x}}}{{\cos x - 1}} = - 2$.
