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Suppose $f\colon [0, +\infty) \to \mathbb{R}$ is a continuous function and $\displaystyle \lim \limits_{x \to +\infty} f(x) = 1$. Find the following limit:

$$\large\displaystyle \lim \limits_{n \to \infty} \int \limits^{2006}_{1385}f(nx)\, \mathrm dx$$

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5 Answers 5

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Let $n\in \Bbb N$ and define $\displaystyle I_n:=\int \limits _{1385}^{2006}f(nx)\,\mathrm dx-621$.

It holds that $\displaystyle I_n =\int \limits _{1385}^{2006}f(nx)-1 \,\mathrm dx$, therefore $\displaystyle |I_n|\leq \int \limits _{1385}^{2006} \left \vert f(nx)-1\right \vert \,\mathrm dx$.

Recall $\lim \limits_{x\to +\infty}f(x)=1 \iff (\forall \delta >0)(\exists \varepsilon >0)(\forall \overline x\in \Bbb R)(\overline x>\varepsilon \implies |f(\overline x)-1|<\delta)$.

Take $\delta >0$. There exists $\varepsilon >0$ such that for all $\overline x>\varepsilon$ it holds that $|f(\overline x)-1|<\delta$. In particular, for large enough $n$, if $x\ge 1385$, then $|f(nx)-1|<\delta$.

It follows that $\displaystyle |I_n|\leq \int \limits _{1385}^{2006}\delta \,\mathrm dx=621\delta$ and $\lim \limits _{n\to +\infty}\left( |I_n|\right)\leq \lim \limits _{n\to +\infty}\left(621\delta\right)=621\delta$.

Since $\delta$ was an arbitrary positive real number, it was proved that $\lim \limits _{n\to +\infty}\left( |I_n|\right)$ is a lower bound of $\{621\delta \colon\delta >0\}$, therefore $\lim \limits _{n\to +\infty}\left( |I_n|\right)\leq \inf \left(\{621\delta \colon \delta >0\} \right)=0$.

Finally $$\displaystyle 0=\lim \limits _{n\to +\infty}\left( |I_n|\right)=\lim \limits _{n\to +\infty}\left( I_n\right)=\lim \limits _{n\to +\infty}\left( \int \limits _{1385}^{2006}f(nx)\,\mathrm dx-621\right),$$ thus $\displaystyle \int \limits _{1385}^{2006}f(nx)\,\mathrm dx =621.$

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  • $\begingroup$ I just realized my change of variable is a bit silly (originally was intended to try to show the integral diverged, which clearly isn't true). I suppose I will adjust it. $\endgroup$
    – Evan
    Sep 6, 2013 at 0:07
  • $\begingroup$ @Evan It's all the same. I got mad when I finished my answer and saw you had already done pretty much the same thing. ^_^ $\endgroup$
    – Git Gud
    Sep 6, 2013 at 0:14
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    $\begingroup$ All good :) Yours has a lot more detail, so it contributes. $\endgroup$
    – Evan
    Sep 6, 2013 at 0:19
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Given $\epsilon > 0$, there is some $M$ for which $|f(x) - 1| \leq \epsilon$ for $x > M$. Then you can show that

$\left| \int_{1385}^{2006} f(nx) dx - (2006-1385)\right| \leq (2006-1385)\epsilon$

for sufficiently large $n$ (when plugging the left bound of the integrand, $1385n > M$). Finally take $\epsilon \to 0$.

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Using the substitution $t=nx$, we get $I_n = \int^{2006}_{1385}f(nx)dx = \frac{1}{n}\int_{1385n}^{2006n} f(t) dt$. Let $I=2006-1385$.

Now let $\epsilon>0$, and choose $L>0$ such that if $t\ge L$, then $-\frac{\epsilon}{I} < f(t)-1 < \frac{\epsilon}{I}$.

Now choose $N\ge \frac{L}{1385}$. Then if $n \ge N$ and $t \in [1385n,2006n]$, we have $-\frac{\epsilon}{I} < f(t)-1 < \frac{\epsilon}{I}$. Integrating over $[1385n,2006n]$ and dividing by $n$ gives $$ -\epsilon < I_n -I < \epsilon $$ It follows that $\lim_n I_n = I$.

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From continuity there is a $\xi_ n \in (1385,2006)$ such that $\int _{1385}^{2006}f(n x) \mathrm{d}x = (2006-1385)f(n \xi _n)$ . Since $k_n= n \xi _n \to \infty$ we have \begin{align}\lim _{n\to \infty}\int _{1385}^{2006}f(n x) \mathrm{d}x &=\lim _{n \to \infty} (2006-1385)f(n \xi _n)\\ &=(2006-1385)\lim _{n \to \infty} f(k_n)=(2006-1385)\end{align}

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Hint The Integral Operator is continuous, so you can swap limit and integral. If you have $f \searrow 1$ (monotonously).

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    $\begingroup$ Not sure it is that fast. If you consider something like $t\mapsto \int \limits_{1385}^tf(nx)dx$, you want the limit in $n$ and not in $t$. $\endgroup$
    – Git Gud
    Sep 5, 2013 at 22:47
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    $\begingroup$ @BertrandR Maybe fix your MathJax before it's too late? ;-) $\endgroup$
    – AlexR
    Sep 5, 2013 at 22:55
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    $\begingroup$ Yes, so for sufficiently large $n$ $|f(nx)| \leq 2$ for all $x \in [1385,2006]$. Dominated (or bounded?) convergence theorem works fine, but not monotone. $\endgroup$
    – Evan
    Sep 5, 2013 at 23:05
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    $\begingroup$ @AlexR: You don't need any convergence theorem. You only need to know that integration preserves order, that is, if $f\le g$ then $\int f \le \int g$. $\endgroup$
    – copper.hat
    Sep 6, 2013 at 0:12
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    $\begingroup$ This hint is misleading (and probably wrong but this is hard to tell since the phrasing is ambiguous). @upvoters Why the upvote? $\endgroup$
    – Did
    Sep 6, 2013 at 8:22

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