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How to compute that $$\lim_{n\to\infty}\int_0^{\pi/2} \sin(x^n)dx$$ $$\lim_{n\to\infty}n^{106}\int_0^{1/n^{2017}} f(x)dx$$ where $f$ is arbitrary integrable function on $[0,1]$

I have tried to swap the limit and integral sign, and get $\int_0^{\pi/2}\displaystyle\lim_{n\to\infty}\sin(x^n)dx$, but there seems no further result can get.. And also the uniform convergence seemingly cannot be justified.

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    $\begingroup$ What is $f$ ? And why did swapping did not work ? $\endgroup$ – Atmos Feb 4 '18 at 14:46
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 17:16
  • $\begingroup$ Do you mean $n^{106}$ or $n^{2016}$? $\endgroup$ – Connor Harris Mar 12 '18 at 17:25
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$$\frac{1}{n}\int_{0}^{(\pi/2)^n}\frac{\sin z}{z^{1-1/n}}\,dz=o\left(\frac{1}{n}\right)+\left(\frac{1}{n}-\frac{1}{n^2}\right)\underbrace{\int_{0}^{(\pi/2)^n}\frac{1-\cos z}{z^{2-1/n}}\,dz}_{O(1)}$$ holds by integration by parts and this gives that the first limit is zero. Try to apply this technique to the second limit, too.

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