Q58 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
The value of $\lim\limits_{x \to 0}\left(\dfrac{8}{x^8}\right)\left[1 - \cos\dfrac{x^2}{2} - \cos\dfrac{x^2}{4} + \cos\dfrac{x^2}{2}\cdot\cos\dfrac{x^2}{4}\right]$ is equal to ...
- A$\dfrac{1}{8}$
- Correct answerB$\dfrac{1}{32}$
- C$\dfrac{1}{16}$
- D$0$
Solution
The given expression inside the limit can be factored as:
$\left(1 - \cos\dfrac{x^2}{2}\right)\left(1 - \cos\dfrac{x^2}{4}\right)$
The limit becomes:
$\lim\limits_{x \to 0} \dfrac{8}{x^8} \left(1 - \cos\dfrac{x^2}{2}\right)\left(1 - \cos\dfrac{x^2}{4}\right)$
Using the standard limit $\lim\limits_{\theta \to 0} \dfrac{1 - \cos \theta}{\theta^2} = \dfrac{1}{2}$, we can rewrite the expression as:
$\lim\limits_{x \to 0} \dfrac{8}{x^8} \left( \dfrac{1 - \cos\dfrac{x^2}{2}}{\left(\dfrac{x^2}{2}\right)^2} \cdot \dfrac{x^4}{4} \right) \left( \dfrac{1 - \cos\dfrac{x^2}{4}}{\left(\dfrac{x^2}{4}\right)^2} \cdot \dfrac{x^4}{16} \right)$
Evaluating the standard limits:
$= \lim\limits_{x \to 0} \dfrac{8}{x^8} \left( \dfrac{1}{2} \cdot \dfrac{x^4}{4} \right) \left( \dfrac{1}{2} \cdot \dfrac{x^4}{16} \right)$
$= \lim\limits_{x \to 0} \dfrac{8}{x^8} \left( \dfrac{x^4}{8} \right) \left( \dfrac{x^4}{32} \right)$
$= \lim\limits_{x \to 0} \dfrac{8}{x^8} \cdot \dfrac{x^8}{256}$
$= \dfrac{8}{256} = \dfrac{1}{32}$
Answer: $\dfrac{1}{32}$
$\left(1 - \cos\dfrac{x^2}{2}\right)\left(1 - \cos\dfrac{x^2}{4}\right)$
The limit becomes:
$\lim\limits_{x \to 0} \dfrac{8}{x^8} \left(1 - \cos\dfrac{x^2}{2}\right)\left(1 - \cos\dfrac{x^2}{4}\right)$
Using the standard limit $\lim\limits_{\theta \to 0} \dfrac{1 - \cos \theta}{\theta^2} = \dfrac{1}{2}$, we can rewrite the expression as:
$\lim\limits_{x \to 0} \dfrac{8}{x^8} \left( \dfrac{1 - \cos\dfrac{x^2}{2}}{\left(\dfrac{x^2}{2}\right)^2} \cdot \dfrac{x^4}{4} \right) \left( \dfrac{1 - \cos\dfrac{x^2}{4}}{\left(\dfrac{x^2}{4}\right)^2} \cdot \dfrac{x^4}{16} \right)$
Evaluating the standard limits:
$= \lim\limits_{x \to 0} \dfrac{8}{x^8} \left( \dfrac{1}{2} \cdot \dfrac{x^4}{4} \right) \left( \dfrac{1}{2} \cdot \dfrac{x^4}{16} \right)$
$= \lim\limits_{x \to 0} \dfrac{8}{x^8} \left( \dfrac{x^4}{8} \right) \left( \dfrac{x^4}{32} \right)$
$= \lim\limits_{x \to 0} \dfrac{8}{x^8} \cdot \dfrac{x^8}{256}$
$= \dfrac{8}{256} = \dfrac{1}{32}$
Answer: $\dfrac{1}{32}$
Hence, the correct answer is option: 'B'.