Explanation

(d)
We know, $\lim_{n \rightarrow \infty}\ x^{2n} = \left\{ \begin{array}{r}
\infty,\ \ \ if\ x^{2} > 1 \\
1,\ \ \ if\ \ x^{2} = 1 \\
0,\ \ \ if\ x^{2} < 1
\end{array} \right.\$
Given, $f(x) = \lim_{n \rightarrow \infty}\frac{\log{(2 + x) - x^{2n}\sin x}}{1 + x^{2n}}$
For $x^{2} = 1,\ \ \ \ \ \ f(x) = \lim_{n \rightarrow \infty}\frac{\log{3 - \sin 1}}{2}$
$= \frac{1}{2}\ (\log{3 - \sin 1})$
For $x^{2} < 1$,
$f(x) = \log{(2 + x)}$
Foe $x^{2} > 1$,
$f(x) = \lim_{n \rightarrow \infty}\frac{\left( \frac{1}{x^{2n}} \right)\log{(2 + x) - \sin x}}{\left( 1 + \frac{1}{x^{2n}} \right)}$
$= - \sin x$
$\therefore\ \ \ \ \ \ \ \ \ \ f(x) = \left\{ \begin{array}{r}
\log{(2 + x),\ \ \ x^{2} < 1} \\
\frac{1}{2}(\log{3 - \sin{1)},\ \ \ \ x = 1} \\
- \sin{x,\ \ \ \ x^{2} > 1}
\end{array} \right.\$
$\therefore\ \lim_{x \rightarrow 1^{-}}\ f(x) = \lim_{h \rightarrow 0}\log(2 + 1 - h)$
=log 3
$\lim_{x \rightarrow 1^{+}}\ f(x) = \lim_{h \rightarrow 0}\lbrack - \sin{(1 + h)}\rbrack$
$= - \sin 1$
It is clear that both limits exist and $\lim_{x \rightarrow 1^{-}}\ f(x) \neq \lim_{x \rightarrow 1^{+}}\ f(x)$