Question

L'hopital's Rule Problem

I don't how my text book gets - 1

Please explain and showing work would help

Answer 1

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

Since
`\cos\frac\pi2=0`, this limit yields the indeterminate form
`\frac00`. By L'Hopital's rule, we have


`\displaystyle\lim_(x\to\pi/2^-)\frac1{-\sin x}`

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