Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\lim_{x\to\infty}\frac{1}{x}\int^x_0f(t)dt=c$?

share|cite|improve this question
Use the l'Hospital rule. – kmitov Feb 16 '14 at 6:57
up vote 4 down vote accepted

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

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

share|cite|improve this answer
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 $0$.

share|cite|improve this answer
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: $$\lim_{x \to \infty}\frac{\int_0^xf(t)dt}{x}=\lim_{x \to \infty}\frac{f(x)}{1}=c$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.