limx→0(1-cos2x3+cosx)/xtan4x=

Correct answer: (D) 2. limx→0⁡((1-cos⁡2x) (3+cos⁡x))/(xtan⁡4x) = limx→0⁡(2sin^2⁡x ⋅(3+cos⁡x))/(x⋅tan⁡4x) = limx→0⁡(2⋅sin⁡xx^2⋅3+cosx)/(tan⁡4x4x) ⋅4 = (2⋅1^2 ⋅(3+1))/(1 ⋅4) =2 .

Q57 / 90 · JEE Main 2015 (04 Apr) · Mathematics

$\lim_{x \to 0} \frac{(1 - cos 2 x) (3 + cos x)}{x tan 4 x} =$

A
$\frac{1}{2}$
B
$4$
C
$3$
Correct answer
D
$2$

Solution

\lim_{x \to 0} \frac{(1 - \cos 2 x) (3 + \cos x)}{x \tan 4 x}

= \lim_{x \to 0} \frac{2 \sin^{2} x \cdot (3 + \cos x)}{x \cdot \tan 4 x}

= \lim_{x \to 0} \frac{2 \cdot {(\frac{\sin x}{x})}^{2} \cdot (3 + \cos x)}{\frac{\tan 4 x}{4 x} \cdot 4}

= \frac{2 \cdot {(1)}^{2} \cdot (3 + 1)}{1 \cdot 4}

= 2

Hence, the correct answer is option: 'D'.

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limx→01-cos2x3+cosxxtan4x=

Solution

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$\lim_{x \to 0} \frac{(1 - cos 2 x) (3 + cos x)}{x tan 4 x} =$