If $\lim_{x \rightarrow 0}\left\{\frac{1}{x^{8}}\left(1-\cos\frac{x^{2}}{2}-\cos\frac{x^{2}}{4}+\cos\frac{x^{2}}{2}\cos\frac{x^{2}}{4}\right)\right\}=2^{-k}$, then the value of $k$ is ______.

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Limits

lim_{x \to 0} {\frac{1}{x ^{8}} (1 - cos \frac{x ^{2}}{2} - cos \frac{x ^{2}}{4} + cos \frac{x ^{2}}{2} cos \frac{x ^{2}}{4})} = 2^{- k}

, then the value of

k

is ______.

Answer: 8

The bracket factors as

(1 - cos \frac{x ^{2}}{2}) (1 - cos \frac{x ^{2}}{4})

. Using

1 - cos θ \approx \frac{1}{2} θ^{2}

for small

θ

, the factors behave like

\frac{1}{2} (\frac{x ^{2}}{2})^{2}

and

\frac{1}{2} (\frac{x ^{2}}{4})^{2}

, so their product is of order

\frac{x ^{4}}{8} \cdot \frac{x ^{4}}{32} = \frac{x ^{8}}{256}

. Dividing by

x^{8}

leaves

\frac{1}{256} = 2^{- 8}

, so

k = 8

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Integer answer

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