Explanation

(a)
This function is continuous at $x = 0$, then
$\lim_{x \rightarrow 0}\frac{\log_{e}{(1 + x^{2}\tan x)}}{\sin x^{3}} = f(0)$
$\Rightarrow \ \ \lim_{x \rightarrow 0}\ \frac{\log_{e}\left\{ 1 + x^{2}\left( x + \frac{x^{3}}{3} + ... \right) \right\}}{x^{3} - \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - ...} = f(0)$
$\Rightarrow \ \lim_{x \rightarrow 0}\ \frac{\log_{e}{(1 + x^{3})}}{x^{3} - \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - ...} = f(0)$
[neglecting higher power of $x$ in $x^{2}\tan x$]
$\Rightarrow \ \ \lim_{x \rightarrow 0}\frac{x^{3} - \frac{x^{6}}{2} + \frac{x^{9}}{3} - \ldots}{x^{3} + \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - ...} = f(0)$
$\Rightarrow \ \ \ \ \ 1 = f(0)$