[tex]\displaystyle\lim_{x\to\pi/2^-}\left(x-\frac\pi2\right)\sec x=\lim_{x\to\pi/2^-}\frac{x-\frac\pi2}{\cos x}[/tex]
Since [tex]\cos\dfrac\pi2=0[/tex], this limit yields the indeterminate form [tex]\dfrac00[/tex]. By L'Hopital's rule, we have
[tex]\displaystyle\lim_{x\to\pi/2^-}\dfrac1{-\sin x}[/tex]
and since [tex]\sin\dfrac\pi2=1[/tex], we end up with -1 as the limit.
