Explanation

(b)
$\lim_{x \rightarrow 0}\frac{e^{x} + \log{(1 + x) - (1 - x)^{- 2}}}{x^{2}}$ $\left\lbrack \frac{0}{0}\ form \right\rbrack$
$= \lim_{x \rightarrow 0}\frac{e^{x} + (1 + x)^{- 1}{- 2(1 - x)^{- 3}}}{2x}$
[by L’ Hospital’s rule]
$= \lim_{x \rightarrow 0}\frac{e^{x} - (1 + x)^{- 2}{- 6(1 - x)^{- 4}}}{2}$
[by L’ Hospital’s rule]
$= \frac{e^{0} - 1 - 6}{2} = - 3$