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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 12. limits4 Marks
Question
$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {1 - cos2x} \right)\left( {3 + \cos x} \right)}}{{x\;tan4x}}$ =
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Answer

d
$\mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\,\tan 4x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x\left( {3 + \cos x} \right)}}{{x \times \frac{{\tan 4x}}{{4x}} \times 4x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}}{{{x^2}}} \times \mathop {\lim }\limits_{x \to 0} \frac{{\left( {3 + \cos x} \right)}}{4} \times \frac{1}{{\mathop {\lim }\limits_{x \to 0} \frac{{\tan 4x}}{{4x}}}}$

$ = 2 \times \frac{4}{4} \times 1$              ($\because $ $\mathop {\lim }\limits_{\theta  \to 0} \frac{{\sin \theta }}{\theta } = 1$ and $\mathop {\lim }\limits_{\theta  \to 0} \frac{{\tan \theta }}{\theta } = 1$)

$=2$

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