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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 12. limits4 Marks
MCQ
$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin (x + a) + \sin (a - x) - 2\sin a}}{{x\sin x}}} \right] = $
  • A
    $\sin a$
  • B
    $\cos a$
  • ✓
    $ - \sin a$
  • D
    $\frac{1}{2}\cos a$
✓

Answer

Correct option: C.
$ - \sin a$
c
(c) $\mathop {\lim }\limits_{x \to 0} \,\,2\,\sin \,a\,.\,\frac{{(\cos x - 1)}}{{x\sin x}}$

$ = - 2\,\sin a\,.\,\frac{{(1 - \cos x)}}{{{x^2}}}\,.\,\left( {\frac{x}{{\sin x}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \,\, - 2\sin a\,.\,\frac{{2\,{{\sin }^2}(x/2)}}{{4\,{{\left( {\frac{x}{2}} \right)}^2}\,\left( {\frac{{\sin x}}{x}} \right)}} = - \sin a$.

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