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).

Answer 1

Correct Answer: 0.24

To solve this problem, we first analyze the limit and the behavior of the logarithmic term as \( t \to \infty \). The integral term is a known standard integral with an oscillatory numerator and decaying denominator. By applying the limits and simplifications to the expression, we find that the value of the entire expression converges to 0.24 when rounded to two decimal places. Thus, the correct answer is 0.24.