Explanation

(c)
Given, $f(x) = x^{3}$
$\therefore\ \ \ \ f(x + h) = (x + h)^{3}$
Now, $f(x) = 3x^{2}$
$\therefore\ \ \ f(x + \theta h) = 3(x + \theta{h)}^{2}$
Given, $\frac{f(x + h) - f(x)}{h} = f^{'}(x + \theta h)$
$\Rightarrow \ \ \ \ \frac{(x + h)^{3} - x^{3}}{h} = 3(x + \theta h)^{2}$
$\Rightarrow \ \ \ \frac{x^{3} + h^{3} + 3xh(x + h) - x^{3}}{h}$
$= 3(x^{2} + \theta^{2}h^{2} + 2x\theta h)$
$\Rightarrow \ \ \ h^{2} + 3x^{2} + 3xh = 3x^{2} + 3{\theta^{2}h}^{2} + 6x\theta h$
$\Rightarrow \ \ \ \ \ h + 3x = 3\theta^{2}h + 6x\theta$
Taking limit on both sides, we get
$\lim_{h \rightarrow 0}(h + 3x) = \lim_{h \rightarrow 0}{(3\theta^{2}h + 6x\theta)}$
$\Rightarrow \ \ \ \ 3x = 6x\ \lim_{h \rightarrow 0}\theta$
$\Rightarrow \ \ \ \lim_{h \rightarrow 0}\theta = \frac{1}{2} = 0.5$