Question

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

• A. 1/8
• B. 0
• C. 1/32
• D. ∞

Answer 1

The correct option is C

1/32

$\text{Put}x=\frac{pi}{2}-hasx\to \frac{pi}{2}h\to 0\therefore \text{Given limit}=\underset{h\to 0}{lim}\frac{1-tan(\frac{\pi }{4}+\frac{h}{2})}{1+tan(\frac{\pi }{4}+\frac{h}{2})}\frac{(1-cosh)}{(2h{)}^3}=\underset{h\to 0}{lim}tan\frac{h}{2}\frac{2si{n}^2\frac{h}{2}}{8h^3}=\underset{h\to 0}{lim}\frac{1}{4}.\frac{tan\frac{h}{2}}{\frac{h}{2}\times 2}\times \left(\frac{sin\frac{h}{2}}{\frac{h}{2}}\right)\times \frac{1}{4}\underset{h\to 0}{lim}\frac{1}{32}\left(\frac{tan\frac{h}{2}}{\frac{h}{2}}\right){\left(\frac{sin\frac{h}{2}}{\frac{h}{2}}\right)}^2=\frac{1}{32}$