Answer
$\dfrac{3}{7}$
Work Step by Step
Consider $f(x)=\lim\limits_{x \to \dfrac{\pi}{2}^{-}}\sec
7 x \cos 3x=\lim\limits_{k \to 0^{+}}\dfrac{\sin 3k}{\sin 7k}=\lim\limits_{k \to 0^{+}} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$
Thus, $f(0)=\dfrac{0}{0}$
This shows an Inderminate form of the limit, so apply L-Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{k \to 0^{+}}\dfrac{3\cos 3k}{7 \cos 7k}=\dfrac{3}{7}$
