MCQ
$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {1 - cos2x} \right)\left( {3 + \cos x} \right)}}{{x\;tan4x}}$ =
- A$ - \frac{1}{4}$
- B$\frac{1}{2}$
- C$1$
- ✓$2$
$ = \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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