Explanation

(b)
Given, $f(x) = \left\{ \begin{array}{r}
\frac{x - 1}{2x^{2} - 7x + 5},\ \ \ \ \ x \neq 1 \\
- \frac{1}{3},\ \ \ \ \ x = 1
\end{array} \right.\$
$f(x) = \left\{ \begin{array}{r}
\frac{1}{2x - 5},\ \ \ \ x \neq 1 \\
- \frac{1}{3},\ \ \ \ \ \ \ \ x = 1
\end{array} \right.\$
$f^{'}(1) = \lim_{h \rightarrow 0}\frac{f(1 + h) - f(1)}{h}$
$= \lim_{h \rightarrow 0}\frac{\frac{1}{2(1 + h) - 5} - \left( - \frac{1}{3} \right)}{h}$
$= \lim_{h \rightarrow 0}{\frac{\frac{1}{2h - 3} + \frac{1}{3}}{h} = \lim_{h \rightarrow 0}\frac{3 + 2h - 3}{3h(2h - 3)} = - \frac{2}{9}}$
$Lf^{'}(1) = \lim_{h \rightarrow 0}\frac{f(1 - h) - f(1)}{- h}$
$= \lim_{h \rightarrow 0}\frac{\frac{1}{2(1 - h) - 5} - \left( - \frac{1}{3} \right)}{- h}$
$= \lim_{h \rightarrow 0} - \frac{2}{3(2h + 3)} = - \frac{2}{9}$
$\therefore\ \ f^{'}(1) = - \frac{2}{9}$