Show that $$\displaystyle\limsup_{x \to \infty} \dfrac{\pi(x)}{x/ \log x} \geqslant 1. $$
I've seen $\displaystyle\lim_{x \to \infty}$ operator, but I haven't seen $\displaystyle\limsup_{x \to \infty}$ or $\displaystyle\liminf_{x \to \infty}$ before. What do they mean exactly? Based on the image from wikipedia:
It seems that we need $\displaystyle\limsup_{x \to \infty}$ and $\displaystyle\liminf_{x \to \infty}$ when a sequence is not strictly increasing or decreasing, but jumping between values. I've not taken analysis, so these terms are not so familiar to me.
$\textbf{Edit:}$ These concepts make sense to me now.
The following was proved in my notes:
For every real number $y \geqslant 2,$
$$\displaystyle\sum\limits_{p \leqslant y} \dfrac{1}{p} > \log(\log y) - 1.$$
This is supposed to be useful in application in this problem, but I am not sure how exactly.
The prime number theorem states that $$\displaystyle\lim_{x \to \infty} \dfrac{\pi(x)}{x/ \log x} = 1,$$ but we didn't prove it in my textbook or notes, so I'd like to refrain from using it if necessary.
This is a classical result of a course in analytic number theory. Chebyshev first showed that $$ c_1\frac{x}{\log(x)}\le \pi(x)\le c_2\frac{x}{\log(x)} $$ for all $x\ge 96098$, with the Chebyshev constants \begin{align*} c_1 & = \log(2^{1/2}3^{1/3}5^{1/5}30^{-1/30})\approx 0.921292022934, \\[0.3cm] c_2 & =\frac{6}{5}c_1\approx 1.10555042752. \\ \end{align*} He proved then the following result:
Proposition(Chebychev): We have $$ 4\log(2)\ge \limsup\limits_{x\rightarrow \infty}\frac{\pi(x)}{x/\log (x)}\ge \liminf\limits_{x\rightarrow \infty}\frac{\pi(x)}{x/\log (x)}\ge \log(2). $$ The upper and lower bounds were finally improved to $1$, implying PNT (Jacques Hadamard and Charles de la Vallée-Poussin). For a reference, e.g., see here.