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As shown in the title, what is the limit $\displaystyle \lim_{x\to\infty}\frac{\int_0^xt-\lfloor t\rfloor \mathrm{d}t}{x}$?

Both numerator and donominator goes to infinity as $x\to\infty$ but the numerator is not always differentiable on $[0,\infty)$, so I think we can't use L'Hopital's Rule? I think the answer should be that the limit DNE but how do we prove that?

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    $\begingroup$ You can directly compute that integral and then use it to compute the limit $\endgroup$
    – Luca.b
    Dec 19 '20 at 14:49
  • $\begingroup$ @Luca.b What do you mean? $\endgroup$ Dec 19 '20 at 14:50
  • $\begingroup$ $\int_0^xt-\lfloor t\rfloor dt=\frac {\lfloor x\rfloor}{2}+\int_0^{x-\lfloor x\rfloor}t dt$ $\endgroup$
    – Luca.b
    Dec 19 '20 at 14:54
  • $\begingroup$ @Luca.b Thank you! $\endgroup$ Dec 19 '20 at 14:58
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Write $n:=\lfloor x\rfloor,\,r:=\{x\}$ so $\int_0^x(t-\lfloor t\rfloor)dt=\tfrac12n+\int_0^rydy=\tfrac12(n+r^2)$, so the limit is squeezed between those of $\frac{n}{2(n+r)}$ and $\frac{n+r}{2(n+r)}=\frac12$, and is $\frac12$.

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  • $\begingroup$ Thank you very much! $\endgroup$ Dec 19 '20 at 14:58
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First, observe that $f(t)=t-\lfloor t\rfloor$ is periodic and bounded: \begin{align} f(t) &= f(t+1)\\ \lvert f(t)\rvert &\le 1\ \forall t \end{align}

Therefore, the primitive of $f$ can be expressed as $$F(x) = ax + B(x)$$ where $a$ is the average value of $f$ over a period, and $B$ is a periodic and bounded function.

$$\therefore\ \lim_{x\to\infty}\frac{\int_0^x f(t)\mathrm{d}t}{x} = \lim_{x\to\infty}\frac{ax+B(x)}{x}$$

Since $B$ is bounded, it is clear that this limit is just equal to $a$, which is given by $$a = \int_0^1 f(t)\mathrm{d}t = \tfrac{1}{2}$$

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