Problem
Compute
$$\lim\limits_{x \to +\infty}\dfrac{\ln x}{\displaystyle \int_0^x \dfrac{|\sin t|}{t}{\rm d}t}.$$
Comment
Maybe, we can solve it by L'Hospital's rule, but there still exists a difficulty here. Though $x \to +\infty$ implies $\ln x \to +\infty$, we do not know the limit of the denominator. How to solve it?
Best Answer
It helps to visualise the functions involved. A cartoonish picture of the integrand is a sequence of bumps of smaller and smaller amplitude. Specifically, for $t$ between $n\pi$ and $(n+1)\pi$, you have $\frac{|\sin t|}{(n+1)\pi}\leq \frac{|\sin t|}{t} \leq \frac{|\sin t|}{n\pi}$. For the first bump, i.e. when $n=0$, we need to take some care: let us replace the upper bound by $\sin t$ in that case.
The integral of $|\sin t|$ over each such interval is $2$ (just integrate $\sin t$ from $0$ to $\pi$, and note that $|\sin t|$ is periodic with period $\pi$ – as always, pictures help), so if we write $f(x)$ for your denominator, then $\sum_{1\leq n\leq x-1}\frac{2}{n\pi}\leq f(x) \leq 2+\sum_{1\leq n\leq x}\frac{2}{n\pi}$. The harmonic sum converges to $\ln x$, so the whole limit is $\pi/2$.