Answer
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\left( {1 - \ln \left( {x + 1} \right)} \right)}}{{\sin x - x}} = \infty $
Work Step by Step
From Table 2 of Section 11.7, 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 $x$
$\ln \left( {1 + x} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{x^n}}}{n} = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \lt 1$ and $x=1$.
Thus, we get
$\sin x - x = - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot$
$1 - \ln \left( {1 + x} \right) = 1 - x + \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} + \dfrac{{{x^4}}}{4} - \cdot\cdot\cdot$
${x^2}\left( {1 - \ln \left( {1 + x} \right)} \right) = {x^2} - {x^3} + \dfrac{{{x^4}}}{2} - \dfrac{{{x^5}}}{3} + \dfrac{{{x^6}}}{4} - \cdot\cdot\cdot$
Hence, the limit becomes
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\left( {1 - \ln \left( {x + 1} \right)} \right)}}{{\sin x - x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - {x^3} + \dfrac{{{x^4}}}{2} - \dfrac{{{x^5}}}{3} + \dfrac{{{x^6}}}{4} - \cdot\cdot\cdot}}{{ - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot}}$
$ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - {x^3} + \dfrac{{{x^4}}}{2} - \dfrac{{{x^5}}}{3} + \dfrac{{{x^6}}}{4} - \cdot\cdot\cdot}}{{ - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - x + \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} + \dfrac{{{x^4}}}{4} - \cdot\cdot\cdot}}{{ - \dfrac{x}{{3!}} + \dfrac{{{x^3}}}{{5!}} - \dfrac{{{x^5}}}{{7!}} + \cdot\cdot\cdot}}$
So, $\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\left( {1 - \ln \left( {x + 1} \right)} \right)}}{{\sin x - x}} = \infty $.
