Answer
\[\lim_{x\rightarrow \left(\frac{π}{2}\right)^-}\cos x\:\sec 5x=\frac{1}{5}\]
Work Step by Step
Let \[l=\lim_{x\rightarrow \left(\frac{π}{2}\right)^-}\cos x\:\sec 5x\]
Which is $0.\infty$ form
\[l=\lim_{x\rightarrow \left(\frac{π}{2}\right)^-}\frac{\cos x}{\cos 5x}\]
Which is $\frac{0}{0}$ form
Using L'Hopital's rule
\[\Rightarrow l=\lim_{x\rightarrow \left(\frac{π}{2}\right)^-}\frac{(\cos x)'}{(\cos 5x)'}\]
\[\Rightarrow l=\lim_{x\rightarrow \left(\frac{π}{2}\right)^-}\frac{-\sin x}{-5\sin 5x}\]
\[\Rightarrow l=\lim_{x\rightarrow \left(\frac{π}{2}\right)^-}\frac{\sin x}{5\sin 5x}\]
\[\Rightarrow l=\frac{1}{5}\]
Hence , \[\lim_{x\rightarrow \left(\frac{π}{2}\right)^-}\cos x\:\sec 5x=\frac{1}{5}\]
