# Evaluate $\lim\limits_{x \to \infty } {\left( {\int_0^{\pi /6} {{{(\sin t)}^x}dt} } \right)^{1/x}}$

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!

## 3 Answers

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

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

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 $1/2$.

If $f(x)$ is continuous on $[a,b]$ and $M=\max \; |f(x)|$, is $M=\lim \limits_{n\to\infty} \left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n}$? We Know that given a continuous function $f:[a,b]\to \Bbb R$ we have, $$\max_{x\in[a,b]}|f(x)|= \lim_{p\to\infty}\left(\int_a^b|f(x)|^pdx\right)^{\frac1p}$$

Hence, $$\lim\limits_{x \to \infty } {\left( {\int_0^{\pi /6} {{{(\sin t)}^x}dt} } \right)^{1/x}} = \color{red}{\max_{x\in[0,\frac\pi6]}\sin x = \sin\frac\pi6 =\frac12.}$$

Given that $\sin x$ is increasing in $[0,\frac\pi6]$.