# lim_(xrarrpi/2) 1-sinx/(x-π/2)?

## A. -2 B. -1 C. 1 D. 0 E. 2

If we set $u = x - \frac{\pi}{2}$ hence $x = u + \frac{\pi}{2}$ we have that

$\lim \left(1 - \left(\sin \frac{u + \frac{\pi}{2}}{u}\right)\right) = \lim \left(1 - \cos \frac{u}{u}\right) = \lim \left(\frac{u - \cos u}{u}\right)$

Because $u \to 0$ and $u - \cos u \to - 1$ we have that the limit is

${\lim}_{u \to 0} \left(\frac{u - \cos u}{u}\right) \to \infty$

Hence none of the given answers is correct.

If you mean the limit $\lim \frac{1 - \sin x}{x - \frac{\pi}{2}}$ then apply D'Hopital law to get

${\lim}_{x \to \frac{\pi}{2}} \frac{\left(1 - \sin x\right) '}{\left(x - \frac{\pi}{2}\right) '} = {\lim}_{x \to \frac{\pi}{2}} \left(- \cos x\right) = - \cos \left(\frac{\pi}{2}\right) = 0$

Hence the correct answer is D. $0$