Suppose $f(x)$ is continuous function $$f:[0,+\infty)\mapsto\mathbb{R}$$ and $$\lim_{x\to \infty}f(x)=1$$ how to find $$\lim_{n\to \infty}\int_{1385}^{2017}f(nx)\,dx $$ Honestly, I have no idea for a start. can someone help me? thanks in advance.
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It seems that the expression under the integral sign converges uniformly to some function. Can you guess what this function is? – Amitai Yuval Nov 25 '17 at 16:51
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1@MarvinF This is totally different: $2017\ne2006$. (haha...) – David C. Ullrich Nov 25 '17 at 17:15
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$\lim\limits_{x\to\infty} f(x) = 1.$ That means no matter how small $\varepsilon>0$ is, if $x$ is big enough, then $1-\varepsilon < f(x) < 1+\varepsilon.$
So let $n$ be so big that when $1385\le x\le 2017$ then $f(nx)$ is big enough to make that inequality hold. Then you have $$ \int_{1385}^{2017} (1-\varepsilon) \,dx \le \int_{1385}^{2017} f(nx)\,dx\le \int_{1385}^{2017} (1+\varepsilon)\,dx. $$ This is true no matter how small $\varepsilon>0$ is.
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Set $a:=1385$ and $b:=2017$; it's clear that when $x\in[a,b]=:D$, then $nx\in[na,nb]=:D_n$ $$ M_n:=\max_{x\in D_n} f(x)\;\;\;,\;\;m_n:=\min_{x\in D_n} f(x). $$ Now by substitution $$ \int_a^bf(nx)\,dx=\frac1n\int_{na}^{nb} f(u)\,du. $$ Then it's clear that $$ m_n(b-a) \le m_n\frac1nn(b-a) \le\frac1n\int_{na}^{nb} f(u)\,du \le M_n\frac1nn(b-a) \le M_n(b-a) $$ Finally, since both $M_n$ and $m_n$ converges to 1, we get the conclusion.
Joe
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