JEE PYQ: Limits Question 12
Question 12 - 2020 (03 Sep Shift 1)
If $\lim_{x \to 0} \left{\frac{1}{x^8}\left(1 - \cos\frac{x^2}{2} - \cos\frac{x^2}{4} + \cos\frac{x^2}{2}\cos\frac{x^2}{4}\right)\right} = 2^{-k}$, then the value of $k$ is ______.
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Answer: 8
Solution
$\lim_{x\to 0} \frac{\left(1-\cos\frac{x^2}{2}\right)\left(1-\cos\frac{x^2}{4}\right)}{x^8} = \lim_{x\to 0}\frac{2\sin^2\frac{x^2}{4}}{x^4/16} \times \frac{2\sin^2\frac{x^2}{8}}{x^4/64} \times \frac{x^4}{16} \times \frac{x^4}{64} \cdot \frac{1}{x^8} = \frac{4}{16 \times 64} = \frac{1}{256} = 2^{-8}$. So $k = 8$.
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