Question

$\underset{y\to \frac{\pi }{2}}{lim}\frac{[1-tan(\frac{x}{2}\left)\right][1-sinx]}{[1+tan(\frac{x}{2}\left)\right][\pi -2x{]}^3}$

• A. $\frac{1}{8}$
• B. $0$
• C. $\frac{1}{32}$
• D. $\infty$

Answer 1

The correct option is C $\frac{1}{32}$

$\underset{y\to \frac{\pi }{2}}{lim}\frac{tan(\frac{\pi }{4}-\frac{x}{2})(1-sinx)}{(\pi -2x{)}^3}Letx=\frac{\pi }{2}+y:y\to 0\Rightarrow \underset{y\to \frac{\pi }{2}}{lim}\frac{tan\left(\frac{-y}{2}\right)(1-cosy)}{(-2y{)}^3}=\underset{y\to \frac{\pi }{2}}{lim}\frac{-tan\frac{y}{2}.2si{n}^2\frac{y}{2}}{(-8)y^3}=\underset{y\to \frac{\pi }{2}}{lim}\frac{1}{32}\frac{tan\frac{y}{2}}{\left(\frac{y}{2}\right)}.[\frac{sin\frac{y}{2}}{\frac{y}{2}}{]}^2=\frac{1}{32}$