Question

L'hopital's Rule Problem

Answer 1

$\displaystyle\lim_{x\to\pi/2^-}\left(x-\frac\pi2\right)\sec x=\lim_{x\to\pi/2^-}\frac{x-\frac\pi2}{\cos x}$

Since $\cos\dfrac\pi2=0$, this limit yields the indeterminate form $\dfrac00$. By L'Hopital's rule, we have

$\displaystyle\lim_{x\to\pi/2^-}\dfrac1{-\sin x}$

and since $\sin\dfrac\pi2=1$, we end up with -1 as the limit.