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STD 11 - 12. limits — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 12. limits1 Mark
MCQ
$\mathop {\lim }\limits_{x \to \pi /6} \left[ {\frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}} \right] = $
  • A
    $\sqrt 3 $
  • ✓
    $1/\sqrt 3 $
  • C
    $ - \sqrt 3 $
  • D
    $ - 1/\sqrt 3 $
✓

Answer

Correct option: B.
$1/\sqrt 3 $
b
(b) Using $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to \pi /6} \frac{{3\cos x + \sqrt 3 \sin x}}{6} = \frac{{3.\frac{{\sqrt 3 }}{2} + \sqrt 3 .\frac{1}{2}}}{6} = \frac{1}{{\sqrt 3 }}$.

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