Question:
The value of
\(\lim\limits_{t→\infin}\left((\log(t^2+\frac{1}{t^2}))^{-1} \int\limits_{1}^{\pi t} \frac{\sin^25x}{x}dx\right)\)
equals ___________ (rounded off to two decimal places).
The value of
\(\lim\limits_{t→\infin}\left((\log(t^2+\frac{1}{t^2}))^{-1} \int\limits_{1}^{\pi t} \frac{\sin^25x}{x}dx\right)\)
equals ___________ (rounded off to two decimal places).
\(\lim\limits_{t→\infin}\left((\log(t^2+\frac{1}{t^2}))^{-1} \int\limits_{1}^{\pi t} \frac{\sin^25x}{x}dx\right)\)
equals ___________ (rounded off to two decimal places).
Updated On: Nov 12, 2024
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Correct Answer: 0.24
Solution and Explanation
The correct answer is 0.24 to 0.26. (approx)
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