# Continuous real-valued function and limit

If $f$ is a continuous real-valued function on $\{x\in \mathbb{R}:x\ge0\}$ and $\lim_{x\to\infty}f(x)=c$, then prove: $$\lim_{x\to\infty}\frac{1}{x}\int^x_0f(t)dt=c.$$

I know that since $\lim_{x\to\infty}f(x) = c$ then $|f(n)-c|<\epsilon$ for $n>N$. But how can I prove that $\lim_{x\to\infty}\frac{1}{x}\int^x_0f(t)dt=c$?

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Use the l'Hospital rule. – kmitov Feb 16 '14 at 6:57

Using the limit of $f$ at $\infty$: For $\epsilon>0,\;\exists A>0,\, |f(x)-c|\le\epsilon$ whenever $x\ge A$ so

$$\left|\frac 1 x\int_0^x f(t)dt-c\right|=\left|\frac 1 x\int_0^x (f(t)-c)dt\right|\le\frac 1x\int_0^A|f(t)-c|dt+\frac1x\int_A^x|f(t)-c|dt\\\le\frac 1x\int_0^A|f(t)-c|dt+\epsilon \le 2\epsilon \;\text{for x large enough}$$

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That was a neat clear way to do it. Thanks a bunch!! – user104235 Feb 16 '14 at 7:02
It's a wrong answer. The continuity of $f$ at $0$ doesn't affect the limit. It's $x\to\infty$, not $x\to0$. – Frank Science Feb 16 '14 at 7:07
@FrankScience I edited my answer and now it's correct;-) – user63181 Feb 16 '14 at 7:11
Sami, is it just me, or was MSE offline for awhile (or otherwise unavailable)? – amWhy Feb 16 '14 at 21:59
Thanks! Me too! It was driving me crazy! – amWhy Feb 16 '14 at 22:08

Fix $\epsilon>0$, and let $x_e$ be such that $|f(x)-c| < \epsilon$ for all $x \geq x_e$.

Then, $\int_0^x f(t) dt = \int_0^{x_e} f(t) dt + \int_{x_e}^x f(t) dt$. The first is just some constant $\gamma$, and the second can be bounded above by $\epsilon (x-x_e)$ and below by $\epsilon(x-x_e)$ since $f$ lies in $[c-\epsilon,c+\epsilon]$ on this interval. Thus, $\frac{1}{x} \int_0^x f(t) dt$ is bounded between $\frac{1}{x} (\gamma-\epsilon(x-x_e))$ and $\frac{1}{x} (\gamma+\epsilon(x-x_e))$. Take the limit as $x \to \infty$ now, and you get the resultant limit lies between $[-\epsilon,\epsilon]$ for arbitrary $\epsilon>0$. Thus, the limit is $0$.

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Thank you very much! – user104235 Feb 16 '14 at 7:02
Is that really squeeze theorem? In fact, what you've shown is that $c-\epsilon\le\liminf\le\limsup\le c+\epsilon$ for arbitrary $\epsilon>0$. – Frank Science Feb 16 '14 at 7:06
You also have a minor glitch. You need to replace the bounds like $\epsilon(x - x_e)$ with $(c - \epsilon)(x - x_e)$ and $(c + \epsilon)(x - x_e)$. Also final limit comes out to be $c$. As noted by Frank Science, the last step consists not of taking a limit, but rather $\limsup$ and $\liminf$ and both these lie between $c - \epsilon$ and $c + \epsilon$. Also it would be better if $f$ is replaced by $f(x) - c$ in the beginning so that we can effectively assume $c$ to be $0$ without losing any generality. – Paramanand Singh Feb 16 '14 at 7:15
you can see Sami Ben Romdhane's answer which uses same idea, but with less hassles. – Paramanand Singh Feb 16 '14 at 7:17

Start by writing the limit as $$\lim_{x \to \infty}\frac{\int_0^xf(t)dt}{x}$$ And use L'Hôpital's rule to get: $$\lim_{x \to \infty}\frac{\int_0^xf(t)dt}{x}=\lim_{x \to \infty}\frac{f(x)}{1}=c$$

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Thank you very much! – user104235 Feb 16 '14 at 7:02
Note that to apply LHR you must show that the integral $\int_{0}^{x}f(t)\,dt$ tends to $\infty$ or $-\infty$ as $x \to \infty$. This can be done (but is not trivial) if $c \neq 0$. If $c = 0$ then you need to think something else. – Paramanand Singh Feb 16 '14 at 7:06
@ParamanandSingh If the integral not tend to $\infty$, the limit is clearly zero, as it is the division of finite by infinite. – LeeNeverGup Feb 16 '14 at 7:09
@ParamanandSingh The general L'Hospital rule doesn't need the condition that the numerator tends to infinity if the denominator tends to positive infinity. – Frank Science Feb 16 '14 at 7:11
Agree Frank, but mostly when students think of LHR, they think of $0/0$ and $\infty/\infty$ so it is better to explicitly state that the more general version is being used. – Paramanand Singh Feb 16 '14 at 7:20