Q81 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics
$\lim\limits_{x \to 0}\dfrac{(1 - \cos 2x)(3 + \cos x)}{x\tan 4x}$ is equal to:
- A$1$
- B$4$
- Correct answerC$2$
- D$3$
Solution
$\lim\limits_{x \to 0}\dfrac{(1 - \cos 2x)(3 + \cos x)}{x\tan 4x}$
Using the trigonometric identity $1 - \cos 2x = 2\sin^2 x$:
$= \lim\limits_{x \to 0}\dfrac{2\sin^2 x(3 + \cos x)}{x\tan 4x}$
Dividing the numerator and the denominator by $x^2$:
$= \lim\limits_{x \to 0}\dfrac{2\left(\dfrac{\sin x}{x}\right)^2(3 + \cos x)}{\dfrac{\tan 4x}{x}}$
$= \lim\limits_{x \to 0}\dfrac{2\left(\dfrac{\sin x}{x}\right)^2(3 + \cos x)}{4\left(\dfrac{\tan 4x}{4x}\right)}$
Using the standard limits $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ and $\lim\limits_{x \to 0}\dfrac{\tan \theta}{\theta} = 1$:
$= \dfrac{2(1)^2(3 + \cos 0)}{4(1)}$
$= \dfrac{2(4)}{4} = 2$
Answer: $2$
Using the trigonometric identity $1 - \cos 2x = 2\sin^2 x$:
$= \lim\limits_{x \to 0}\dfrac{2\sin^2 x(3 + \cos x)}{x\tan 4x}$
Dividing the numerator and the denominator by $x^2$:
$= \lim\limits_{x \to 0}\dfrac{2\left(\dfrac{\sin x}{x}\right)^2(3 + \cos x)}{\dfrac{\tan 4x}{x}}$
$= \lim\limits_{x \to 0}\dfrac{2\left(\dfrac{\sin x}{x}\right)^2(3 + \cos x)}{4\left(\dfrac{\tan 4x}{4x}\right)}$
Using the standard limits $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ and $\lim\limits_{x \to 0}\dfrac{\tan \theta}{\theta} = 1$:
$= \dfrac{2(1)^2(3 + \cos 0)}{4(1)}$
$= \dfrac{2(4)}{4} = 2$
Answer: $2$
Hence, the correct answer is option: 'C'.