Question

$\underset{x\to \pi /2}{lim}\frac{1-\sin x}{{(x-\frac{\pi }{2})}^2}$ equals

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{2}$

Let, $x=\frac{\pi }{2}-h$, So as $x\to \frac{\pi }{2},h\to 0$

$\therefore \underset{x\to \pi /2}{lim}\frac{1-\sin x}{{(x-\frac{\pi }{2})}^2}=\underset{h\to 0}{lim}\frac{1-\sin (\frac{\pi }{2}-h)}{h^2}$

$=\underset{h\to 0}{lim}\frac{1-\cos h}{h^2}=\underset{h\to 0}{lim}\frac{2{\sin }^2\frac{h}{2}}{h^2}=\frac{1}{2}\underset{h\to 0}{lim}{\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2=\frac{1}{2}$