Question

$ \displaystyle\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin \, x}{\cos \, x}$ is equal to

• A. 0
• B. -1
• C. 1
• D. does not exist

Answer 1

Answer: (A)

Since $ \displaystyle\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin \, x}{\cos \, x}$
$= \displaystyle\lim_{y \to 0}\left[\frac{1-\sin\left(\frac{\pi}{2}-y\right)}{\cos\left(\frac{\pi}{2}-y\right)}\right]$
(taking $\frac{\pi}{2} - x = y$)
$ = \displaystyle\lim_{y \to 0} \frac{1 - \cos\, y}{\sin\, y}$
$= \displaystyle\lim_{y \to 0} \frac{2 \, \sin^2 \frac{y}{2}}{2 \, \sin \frac{y}{2} \cos \frac{y}{2}}$
$ = \displaystyle\lim_{y \to 0} \: \tan \frac{y}{2} = 0$