MCQ
$\mathop {\lim }\limits_{x \to 1} (1 - x)\tan \left( {\frac{{\pi x}}{2}} \right) = $
- A$\frac{\pi }{2}$
- B$\pi $
- ✓$\frac{2}{\pi }$
- D$0$
Put $1 - x = y$ as $x \to 1,\,\,y \to 0$
Thus $\mathop {\lim }\limits_{y \to 0} \,\,y\tan \frac{{\pi \,(1 - y)}}{2} = \mathop {\lim }\limits_{y \to 0} \,\,\frac{2}{\pi }.\frac{{\left( {\frac{{\pi y}}{2}} \right)}}{{\tan \,\left( {\frac{{\pi y}}{2}} \right)}} $
$= \frac{2}{\pi } \times 1 = \frac{2}{\pi }$.
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