Show that $\limsup_{x \to \infty} \frac{\pi(x)}{x/ \log x} \geqslant 1. $ 
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.
 A: 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.
A: Use $$\sum_{p \leq x} \frac{1}{p}= \frac{\pi(x)}{x}+ \int_2^x \frac{\pi(y)}{y^2} dy,$$(this can be verified easily,e.g. Show that $\sum\limits_{p \leqslant x}1/p = \frac{\pi(x)}{x} + \int_2^x \frac{\pi(u)}{u^2} du.$)
and prove by contradiction.
By some simple calculation, when $x$ is sufficiently large, RHS of the above equality becomes strictly smaller than $log log x-1$, contradicting to the given fact
$$\sum_{p \leq x} \frac{1}{p} >log log x −1.$$
