Explanation

(d)
$\lim_{\theta \rightarrow 0}\frac{4\left( \tan{\theta - 2\theta^{2}\tan\theta} \right)}{\left( 1 - \cos{2\theta} \right)}$
$= \ \frac{4\left( \theta\sec^{2}{\theta +}\tan{\theta - 4\theta\tan\theta - 2\theta^{2}\sec^{2}\theta} \right)}{2\sin{2\theta}}$
[using L’ Hospital’s rule]
$= \lim_{\theta \rightarrow 0}\$
$\frac{\left\lbrack 4\left( \begin{array}{r}
\sec^{2}{\theta +}2\theta{\sec^{2}\theta\tan}{\theta + \sec^{2}\theta - 4\tan\theta - \ } \\
4\theta\sec^{2}\theta - 4\theta\sec^{2}\theta - 4\theta^{2}\sec^{2}\theta\tan\theta
\end{array} \right) \right\rbrack}{4\cos{2\theta}}$
[using L’ Hospital’s rule]
$= \frac{4(1 + 0 + 1)}{4} = 2$