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Let $f$ be a continuous function in $\left[0,\infty\right]$ such that $\displaystyle \lim_{x\to\infty}f(x)=a$ Prove $\displaystyle \lim_{x\to\infty}\frac{1}{x}\int_{0}^{x}f(t)dt=a$.

Well, i made this:

As $f$ is a continuos function in $\left[0,\infty\right]$ by the fundamental theorem of calculus I exists a $F(x)$ such that $f(x)=F'(x)$ and by the FTC II we have $\int_{0}^{x}f(t)dt=f(x)-f(0)$, Then:

$\displaystyle \lim_{x\rightarrow\infty}\frac{1}{x}\int_{0}^{x}f(t)dt=\lim_{x\rightarrow\infty}\frac{1}{x}(f(x)-f(0))=0-0=0\neq a $

Can someone help?

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  • $\begingroup$ What is $[0,\infty]$? Is $\infty$ in the domain of $f$? $\endgroup$ – Kenny Lau May 15 '17 at 15:11
  • $\begingroup$ @KennyLau Is an interval... $\endgroup$ – Bvss12 May 15 '17 at 15:12
  • $\begingroup$ $\int_0^x f(t)dt = F(x)-F(0)$, not $f(x)-f(0)$ $\endgroup$ – Fan May 15 '17 at 15:14
  • $\begingroup$ I think you should write $\infty)$ instead of $\infty]$ $\endgroup$ – Jaideep Khare May 15 '17 at 15:15
  • $\begingroup$ Sure @JaideepKhare is true, sorry. $\endgroup$ – Bvss12 May 15 '17 at 15:21
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You are trying to compute an indeterminate form. Just use L'Hôpital's rule.

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  • $\begingroup$ i don't see the indermination, can you explain where is the indetermination? I know how solve that limit with L'hopital but i want know where is the indetermination $\endgroup$ – Bvss12 May 15 '17 at 15:49
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    $\begingroup$ Unless $a=0$, the limit that you want to determine os of the type $\frac\infty\infty$. $\endgroup$ – José Carlos Santos May 15 '17 at 17:26
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Hint: use L'Hopital's Rule together with the first part of the Fundamental Theorem of Calculus.

EDIT: in your tentative solution you say that $0\neq a$ and let's assume without loss that $a > 0$. Then there is $x_0$ such that for $x \ge x_0$ $|f(x) - a| < \frac{a}{2}$, which implies $f(x) \ge \frac{a}{2}.$ But then $$\int_0^{\infty}f(x)\,dx \ge \int_0^{x_0}f(x)\,dx + \frac{a}{2}\int_{x_0}^{\infty}1\,dx = \infty.$$ This shows that the numerator blows up to $+\infty$. The denominator is $x$, which obviously converges to $+\infty$ as well and hence you have an indeterminate form.

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  • $\begingroup$ Giovanni, i don't see the indermination, can you explain where is the indetermination? I know how solve that limit with L'hopital but i want know where is the indetermination $\endgroup$ – Bvss12 May 15 '17 at 15:48
  • $\begingroup$ @Bvss12: good question, I've edited my post with the answer. $\endgroup$ – Giovanni May 15 '17 at 16:18
  • $\begingroup$ You don't need to worry about the numerator. If the denominator $\to \infty,$ you can apply LHR. $\endgroup$ – zhw. May 15 '17 at 19:33
  • $\begingroup$ @zhw.: you are of course right, thanks for your comment! $\endgroup$ – Giovanni May 15 '17 at 20:19
  • $\begingroup$ @Bvss12: as zhw. remarked, the assumption that the numerator diverges is not used in the proof in the case that the denominator $\to \infty$. Moreover, thanks to this insightful remark, we can also notice that the assumption $a \neq 0$ is not needed. $\endgroup$ – Giovanni May 15 '17 at 20:19
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For each $x>0$ and $c>0$ with $0<c<x$, $$\begin{align*}\left|\frac1x\int_0^x f(t)\,dt - a\right|&=\left|\frac1x\int_0^x f(t)-a\,dt\right|\\ &=\left|\frac1x\int_0^c f(t)-a\,dt+\frac1x\int_c^x f(t)-a\,dt\right|\\ &\leq\left|\frac1x\int_0^c f(t)-a\,dt\right|+\left|\frac1x\int_c^x f(t)-a\,dt\right|\\ &\leq \frac{c(\max_{t\in[0,c]}|f(t)|+|a|)}{x}+\frac{x-c}{x}\sup_{t\geq c}|f(t)-a|. \end{align*}\\$$

Given $\varepsilon>0$, let $c>0$ be such that $t\geq c$ implies $|f(t)-a|<\frac12\varepsilon$. Let $x_0>c$ be sufficiently large that $\frac{c(\max_{t\in[0,c]}|f(t)+|a|)}{x_0}<\frac12\varepsilon$. Then for all $x\geq x_0$, $\left|\frac1{x}\int_0^x f(t)\,dt-a\right|<\varepsilon.$

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  • $\begingroup$ Jonas, other people in the post say, the limit have an indetermination, can you explain me where is the indetermination? $\endgroup$ – Bvss12 May 15 '17 at 15:49
  • $\begingroup$ @Bvss12 You don't see the indetermination because in the second-to-last passage you forgot the $\frac1x$. Also, you confused capital $F$ with $f$ in FTC II. $\endgroup$ – user228113 May 15 '17 at 15:59
  • $\begingroup$ @Bvss12: $\lim\limits_{x\to\infty}\dfrac{\int_0^x f(t)\,dt}{x}$ is an indeterminate form because both numerator and denominator go to $\infty$ as $x\to\infty$. $\endgroup$ – Jonas Meyer May 15 '17 at 16:12
  • $\begingroup$ @JonasMeyer I see very clear denominator go to $\infty$ but, i dont see $\int_0^x f(t)\,dt$ because go to $\infty$ $\endgroup$ – Bvss12 May 15 '17 at 16:14
  • $\begingroup$ @ $\int_0^x f(t)\,dt$= $F(x)-F(0)$ no?, i dont see $F(x)-F(0)$ go to $\infty$ $\endgroup$ – Bvss12 May 15 '17 at 16:16
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Since $f(x)\to a$ at infinity, there exists $x_0$ such that $\forall x\ge x_0$ we have $a-\varepsilon\le f(x)\le a+\varepsilon$

We have $\displaystyle \frac 1x\int_0^x f(t)dt=\frac 1x\underbrace{\int_0^{x_0} f(t)dt}_{\text{constant}=C}+\frac 1x\int_{x_0}^x f(t)dt$

So $\displaystyle\frac{(a-\varepsilon)(x-x_0)+C}x\le \frac 1x\int_0^x f(t)dt\le \frac{(a+\varepsilon)(x-x_0)+C}x$

And the conclusion is immediate.

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You can argue this, as known

$$ \int_{0}^{x} f(x)dx = \int_{0}^{x_0} f(x)dx + \int_{x_0}^{x} f(x)dx $$

Take $\varepsilon > 0$.

$\hspace{1cm}$Once $\lim\limits_{x \rightarrow \infty} f(x) = a$, given $\frac{\varepsilon}{2}>0$ exists $x_0$ such that $|f(x) - a| < \frac{\varepsilon}{2}$, $\forall$ $x$ $>$ $x_0$. Let $M = \sup\limits_{x \in [0,x_0]} {|f(x) - a|}$.

So for all $x > \max\{{x_0, \frac{2M}{\varepsilon}}\}$, we have

\begin{align*}\left| \frac{1}{x} \int_{0}^{x} f(x)dx - a \right| &= \left| \frac{1}{x}\int_{0}^{x_0} f(x)dx + \frac{1}{x}\int_{x_0}^{x} f(x)dx - a \right|\\ &= \left| \frac{1}{x}\int_{0}^{x_0} f(x)dx + \frac{1}{x}\int_{x_0}^{x} f(x)dx - \frac{x}{x}a \right|\\ &= \left| \frac{1}{x}\int_{0}^{x_0} f(x) - a \hspace{0.1cm}dx + \frac{1}{x}\int_{x_0}^{x} f(x) - a \hspace{0.1cm}dx \right|\\ &\leq \left| \frac{1}{x}\int_{0}^{x_0} f(x) - a \hspace{0.1cm}dx \right| + \left| \frac{1}{x}\int_{x_0}^{x} f(x) - a \hspace{0.1cm}dx \right|\\ &\leq \frac{1}{x} M + \frac{\varepsilon}{2} \left|\frac{x-x_0}{x} \right|\\ &\leq \frac{\varepsilon}{2M} M + \frac{\varepsilon}{2} \\ & \leq \varepsilon \\ \end{align*}

So $\lim\limits_{x \rightarrow \infty} \frac{1}{x}{\int_{0}^{\infty} f(x)dx} = a$

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