Suppose $x_n$ and $y_n$ are a sequence of real numbers such that $$\limsup_{n\rightarrow \infty} \frac{x_n}{y_n} = 1 \quad \mathrm{and} \quad \liminf_{n\rightarrow \infty} \frac{x_n}{y_n} = -1.$$
Intuitively this says that as $n \rightarrow \infty$ the quotient $x_n/y_n$ remains within the the interval $(-1, 1)$. Of course this is only a limiting statement, so that for any fixed $n$ the quotient may take values outside $(-1, 1)$, but we expect this to happen less often as $n$ gets larger.
However I am following a proof (found here: Asymptotics of Brownian motion) that says the above implies that for any $\epsilon > 0$ we have $$x_n \geq (1-\epsilon)y_n \quad \mathrm{and} \quad x_n \leq -(1-\epsilon)y_n.$$
But $\limsup$ and $\liminf$ are only limiting statements so how can the above be true for a particular value $n$ or even all $n$?
The cited proof doesn't state that $x_n\geq (1-\varepsilon) y_n$.
Instead, it follows this reasoning:
As $\text{lim sup}\space \dfrac{x_n}{y_n}=1$, there exists a sequence $t_n$ such that $\dfrac{x_{t_n}}{y_{t_n}}\to 1$.
Then, there exists $n_0\in\mathbb{N}$ such that, if $n\geq n_0$, $\dfrac{x_{t_n}}{y_{t_n}}\geq (1-\varepsilon)\iff x_{t_n}\geq (1-\varepsilon) y_{t_n}.$
As the author of the original text is only interested in the asymptotic behaviour of the Brownian Motion (that is, when $n\to\infty$), it is not relevant this only occurs after $n_0$.