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Is there a continuous function $f : ]1, \infty[ \to \Bbb R_{>0}$ such that $f(x) \to 0$ when $x \to +\infty$ but $$\lim_{x \to +\infty} \dfrac 1 x \int_1^x f \;\neq\; 0$$ (either the limit doesn't exist but a better example for me would be when it exists and is $>0$)?

I tried something like $f(t) = \dfrac 1 {\log\big(\log(t)\big)}$ but I'm not sure how to handle the integral. The problem seems to be when $f$ doesn't go to $0$ sufficient quickly.

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    There is none. The proof is a simple exercise on the X-epsilon definition of limit. – 2017-02-27
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    @Did : I tried that, but I'm stuck if $f$ takes very large values when $x \to 1$, for instance $\lim_{x \to 1} f(x) = +\infty$ "very quickly". – 2017-02-27
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    This is irrelevant since you know by hypothesis, say, that $$\int_1^2f(t)dt$$ is finite. – 2017-02-27
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    If $f(x)$ is continuous on $[1,\infty)$ then $f(1)$ has to be finite thus you cannot have $f\to \infty$ as $x\to 1$. If you instead want to consider continuous functions on $(1,\infty)$, for which it can happen, then note that if you are to ask for the limit of something then that something has to exist in the first place. For example it does not make much sense to ask what the limit of $\frac{1}{x}\int_1^x \frac{{\rm d}t}{t-1}$ is as the integral does not exist for any $x$. – 2017-02-27

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No, there is not.

Lets take $\varepsilon$ and prove that there exists $N$ s.t. for $x>N$ the mean $M(n) =\dfrac 1 x \int_1^x f(x)\,dx \in [-\varepsilon,\varepsilon]$

Suppose that for $x> Q~|f(x)| < \delta $. Then

$$M(2Q) = \frac12M(Q)+\frac1{2Q} \int_1^{2M} f(x)dx\le \frac12M(Q)+\frac1{2Q} \int_1^{2Q} \delta \,dx = \frac12M(Q)+\delta$$ Similarly $$M(4Q) \le \frac12M(2Q)+\delta \le \frac14M(Q)+(1+\frac12)\delta$$ and so on. So we will finally find the power $q$ such that $M(2^qQ)<\varepsilon$.

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Writing the expression as

$$\tag 1 \frac{\int_1^x f(t)\,dt}{x}$$

we can apply L'Hopital (since the denominator $\to \infty$). The ratio of derivatives is simply $f(x)/1=f(x),$ which is given to $\to 0.$ Therefore the limit of $(1)$ is $0.$

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No.

If $f(x)\to 0$, then you can find an $x_0$ such that $f(x)<\varepsilon/2$ everywhere to the right of $x_0$, and then no matter how large $\int_1^{x_0} f(t)\,dt$ is, you can always push $\frac1x \int_1^x f(t)\,dt$ below $\varepsilon$ by choosing $x$ far enough to the right of $x_0$.

Of course you do need to assume that, say, $\int_1^2 f(t)\,dt$ exists (and is finite) at all. For example, if $f(t)=\frac{1}{t-1}$, then $\frac1x\int_1^xf(t)\,dt$ never exists (or is always infinite, depending on your conventions).

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    Thank you for your answer. But is there no problem if $f$ takes very large values when $x \to 1$, for instance $\lim_{x \to 1} f(x) = +\infty$ "very quickly"? I will think about it anyway. – 2017-02-27
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    @Alphonse: I've added a paragraph discussing that to the answer. – 2017-02-27