Question

$\underset{x\to \frac{\pi }{2}}{\lim }\frac{1-\sin x}{\cos x}$ is equal to

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

Answer 1

Answer: (A)

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