Answer
$-2$
Work Step by Step
Consider: $\lim\limits_{\theta \to \dfrac{\pi}{2}}f(x)=\lim\limits_{ \theta \to \dfrac{\pi}{2}}\dfrac{2 \theta -\pi}{\cos (2\pi-\theta)}$
Need to check that the limit has an indeterminate form.
Thus, $f(\dfrac{\pi}{2})=\dfrac{2 (\dfrac{\pi}{2}) -\pi}{\cos (2\pi-(\dfrac{\pi}{2}))}=\dfrac{0}{0}$
Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
This implies:
$\lim\limits_{\theta \to \dfrac{\pi}{2}}\dfrac{\cos x -1}{3x^2}=\dfrac{0}{0} $
Again apply L-Hospital's rule:
we get $\lim\limits_{\theta \to \dfrac{\pi}{2}}\dfrac{2}{\sin (2\pi -\theta)}=\dfrac{2}{\sin (2\pi -\dfrac{\pi}{2})}=-2 $
