# Limit of $\lim_{x\to\infty}\frac{\int_0^xt-\lfloor t\rfloor dt}{x}$

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?

• You can directly compute that integral and then use it to compute the limit Dec 19 '20 at 14:49
• @Luca.b What do you mean? Dec 19 '20 at 14:50
• $\int_0^xt-\lfloor t\rfloor dt=\frac {\lfloor x\rfloor}{2}+\int_0^{x-\lfloor x\rfloor}t dt$ Dec 19 '20 at 14:54
• @Luca.b Thank you! Dec 19 '20 at 14:58

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 $$\frac12$$.
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 $$a = \int_0^1 f(t)\mathrm{d}t = \tfrac{1}{2}$$