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12. limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS12. limits1 Mark
MCQ
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {1 - \tan \left( {\frac{x}{2}} \right)} \right]\,[1 - \sin x]}}{{\left[ {1 + \tan \left( {\frac{x}{2}} \right)} \right]\,{{[\pi - 2x]}^3}}}$ is
  • A
    $\frac{1}{8}$
  • B
    $0$
  • ✓
    $\frac{1}{{32}}$
  • D
    $\infty $
✓

Answer

Correct option: C.
$\frac{1}{{32}}$
c
(c) $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right)\,(1 - \sin x)}}{{{{(\pi - 2x)}^3}}}$
Let $x = \frac{\pi }{2} + y:y \to 0$ ==> $\mathop {\lim }\limits_{y \to 0} \frac{{\tan \left( {\frac{{ - y}}{2}} \right)\,(1 - \cos y)}}{{{{( - 2y)}^3}}}$
= $\mathop {\lim }\limits_{y \to 0} \frac{{ - \tan \frac{y}{2}.2{{\sin }^2}\frac{y}{2}}}{{( - 8){y^3}}}$ $ = \mathop {\lim }\limits_{y \to 0} \frac{1}{{32}}\frac{{\tan \frac{y}{2}}}{{\left( {\frac{y}{2}} \right)}}.{\left[ {\frac{{\sin \frac{y}{2}}}{{\frac{y}{2}}}} \right]^2} = \frac{1}{{32}}$.

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