Answer
$-\dfrac{5}{3}$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to -\dfrac{\pi}{2}}\sec(3x) \cos (5x)=\lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{\cos 5x}{\cos 3x}$.
But $\lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{\cos 5x}{\cos 3x}=\dfrac{\infty}{\infty}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{\cos 5x}{\cos 3x}=\lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{-5 \sin (5x)}{-3 \sin (3x)} \\ =\dfrac{5}{3} \times \lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{\sin (5\pi/2)}{\sin (3 \pi/2)} \\=\dfrac{5}{3} \times \dfrac{1}{-1} \\=-\dfrac{5}{3}$