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Limits Continuity and Differentiability question

2013 · Shift 0 · Q29

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JEE Main
Mathematics
Limits Continuity and Differentiability
2013 · Shift 0 · Q29

Limits Continuity and Differentiability question

2013 · Shift 0 · Q29

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

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

is equal to

A
$- {1 \over 4}$
B
${1 \over 2}$
C
1
D
2

View written solutionFree

Correct answer: D

We need to evaluate

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

Use standard small-angle limits / approximations near $x=0$ :

1-\cos 2x \sim \frac{(2x)^2}{2}=2x^2,

\cos x \to 1 \implies 3+\cos x \to 4,

\tan 4x \sim 4x.

So,

x\tan 4x \sim x\cdot 4x=4x^2.

Substitute these into the expression:

L=\lim_{x\to 0}\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x} \sim \frac{(2x^2)(4)}{4x^2}=2.

Hence,

L=2.

Check options:

A: $-\frac14$ ❌
B: $\frac12$ ❌
C: $1$ ❌
D: $2$ ✅

Therefore, the correct answer is D.

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