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It is given that the following limit $\mathop {\lim }\limits_{x \to \infty } {\left( {\int\limits_0^{\pi /6} {{{(\sin t)}^x}dt} } \right)^{1/x}}$ exists. Evaluate the limit.

I've tried tackling this problem but I can't seem to get started. Any hint is appreciated, thanks!

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2 Answers 2

Note that the desired integral is the $L^x$ norm of $\sin t$ on the interval $[0,\pi/6]$. It is well known that the $L^x$ norm of a function converges to the $L^\infty$ norm as $x\rightarrow \infty$. (For example, this is an exercise in Folland's book on real analysis.) It is easy to see that the $L^\infty$ norm of $\sin t$ on this interval is $1/2$.

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Note that for every $x>\frac{6}{\pi}$, when $\frac{\pi}{6}-\frac{1}{x}\le t\le \frac{\pi}{6}$, $0<\sin\left(\frac{\pi}{6}-\frac{1}{x}\right) \le \sin t\le \frac{1}{2}$. It follows that $$\sin\left(\frac{\pi}{6}-\frac{1}{x}\right)\left(\frac{1}{x}\right)^{\frac{1}{x}}\le\left(\int_0^{\frac{\pi}{6}}(\sin t)^x dt\right)^{\frac{1}{x}}\le \frac{1}{2}\left(\frac{\pi}{6}\right)^{\frac{1}{x}}.$$ Letting $x\to\infty$, it follows that

$$\lim_{x\to\infty}\left(\int_0^{\frac{\pi}{6}}(\sin t)^x dt\right)^{\frac{1}{x}}=\frac{1}{2}.$$

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