Explanation

(a)
Given, $f(x) = x - |x - x^{2}|$
At $x = 1$, $f(1) = 1 - |1 - 1| = 1$
$\lim_{x \rightarrow 1^{- 1}}{f(x)} = \lim_{h \rightarrow 0}{\lbrack(1 - h) - |(1 - h){- (1 - h)}^{2}|\rbrack}$
$= \lim_{h \rightarrow 0}\ \left\lbrack (1 - h) - \left| h - h^{2} \right| \right\rbrack = 1$
$\lim_{x \rightarrow 1^{+}}{f(x)} = \lim_{h \rightarrow 0}\lbrack(1 + h) - |(1 + h) - (1 + h)^{2}|\rbrack$
$= \lim_{h \rightarrow 0}\ \ \left\lbrack 1 + h - \left| - h^{2} - h \right| \right\rbrack = 1$
$\because\ \ \lim_{x \rightarrow 1^{- 1}}{f(x)} = \lim_{x \rightarrow 1^{+}} = f(1)\$