Approximation in a equality Let $X_1,X_2$ be independent distributed with cdfs $F_1(x),F_2(x)$, so that 
$$\overline F_i(x) := 1 - F_i(x) = x^{-\alpha}L_i(x),\ \alpha \geq 0 \>,$$
where
$L_i(x)$ is a slowly varying function, that is, it satisfies $L(tx)/L(x) \to 1$ as $x \to \infty$ for all $t > 0$.
For $0<\delta<\frac{1}{2}$ I have the following line:
$\overline{F_{1}}((1-\delta)x)+\overline{F_{2}}((1-\delta)x)+\overline{F_{1}}(\delta x)\overline{F_{2}}(\delta x)
=\left(\overline{F_{1}}((1-\delta)x)+\overline{F_{2}}((1-\delta)x)\right)\left(1+o(1)\right).$
Can someone explain why we have $o(1)$ as $x \rightarrow \infty$ in the second line of the equation?
 A: Thanks @cardinal for the definition.
Simplifying on both sides of the $=$ sign, we get
$$ \frac{\bar{F}_1(\delta x) \bar{F}_2(\delta x)}{\bar{F}_1((1-\delta) x) + \bar{F}_2((1-\delta) x)} = o(1), $$
which means that it tends to 0 (as $x \rightarrow \infty$ since you specified it).
Is your question about the meaning of $o(1)$ or how to prove this inequality? If it is the first case, the small $o$ Landau notation $f(x) = o(g(x))$, $x\to a$ sometimes said "$f$ is negligible compared to $g$ in the neighborhood of $a$", means that the function $f$ is such that 
$$ \lim_{x\to a} \frac{f(x)}{g(x)} = 0, $$
so $o(1)$ means that $f$ tends to 0.
If your question is actually the second one, I feel we miss some information, like can $\delta$ be greater than 1 (in which case the limit is trivially true)?
EDIT: with the new assumptions you can prove it as follows:
$$ \frac{\bar{F}_1(\delta x) \bar{F}_2(\delta x)}{\bar{F}_1((1-\delta) x) + \bar{F}_2((1-\delta) x)} = \frac{(\delta x)^{-2\alpha}L_1(\delta x)L_2(\delta x)}{((1 - \delta)x)^{-\alpha} (L_1((1-\delta) x) + L_2((1-\delta) x))}. $$
Dividing numerator and denominator by $L_1(x)L_2(x)$ and taking the limit, the whole things is equivalent to
EDIT:
$$\frac{(\delta x)^{-2\alpha}}{((1 - \delta)x)^{-\alpha}(\frac{1}{L_2(x)} + \frac{1}{L_1(x)})}.$$
By hypothesis $L_i(x) = o(x^\alpha)$, in the limit $\frac{1}{L_i(x)} > x^{-\alpha}$ so the denominator is bounded and the ration tends to 0.
