Show that $\mathbb{P}(x\leq X\leq x+\Delta x)=f_{X}(x)\Delta x+o(\Delta x)$.

Question

I am reading A Course in Queuing Theory by Moshe Haviv. In the fourth page, he says the following.

Denote by $o(x)$ any function of $x$ such that $\lim_{x\rightarrow 0}\frac{o(x)}{x}=0.$ Let $X$ be a random variable, with density function $f_{X}(x)$ and with tail function $\overline{F}_{X}(x):=\mathbb{P}(X\geq x)$. Then, we have $$\dfrac{\mathbb{P}(x\leq X\leq x+\Delta x)}{\mathbb{P}(X\geq x)}=\dfrac{f_{X}(x)\Delta x+o(\Delta x)}{\overline{F}_{X}(x)}=\dfrac{f_{X}(x)\Delta x}{\overline{F}_{X}(x)}+o(\Delta x).$$

Firstly, I have some partial answer for the equation $$\mathbb{P}(x\leq X\leq x+\Delta x)=f_{X}(x)\Delta x+o(\Delta x).$$ I think that he sort of used Taylor Theorem. By definition, we have $$\mathbb{P}(x\leq X\leq x+\Delta x)=\int_{x}^{x+\Delta x}f_{X}(t)dt=F_{X}(x+\Delta x)-F_{X}(x).$$ We Taylor expand $F_{X}(x+\Delta x)$ around $a=x$ so that $$F_{X}(x+\Delta x)=F_{X}(x)+F'_{X}(x)(x+\Delta x-x)+o(|x+\Delta x-x|^{2}),$$ but following this procedure, we have $$F_{X}(x+\Delta x)=F_{X}(x)+f_{X}(x)\Delta x+o((\Delta x)^{2}),$$ so that $$\mathbb{P}(x\leq X\leq x+\Delta x)=F_{X}(x+\Delta x)-F_{X}(x)=f_{X}(x)\Delta x+o((\Delta x)^{2}).$$ But in his equation, it is $o(\Delta x)$. Am I missing anything?

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I don't quite understand how he gets the second equality. Does he sort of assume that $o(\Delta x)$ is pretty small so $\frac{o(\Delta x)}{\overline{F}_{X}(x)}\approx o(\Delta x)$? But the little o notation only tells us that $o(\Delta x)$ goes to 0 faster than $\Delta x$ when $\Delta x$ goes to zero, right?

Thanks for help!

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Edit 1: About the Second Equality.

Yeah it turned out that I was confused about the little o notation. But I understand it now. So, $$\dfrac{o(\Delta x)}{\overline{F}_{X}(x)}\rightarrow 0\ \ \text{as}\ \ \Delta x\rightarrow 0,$$ which means that $\frac{o(\Delta x)}{\overline{F}_{X}(x)}=o(\Delta x)$ by definition.

Answer 1

It follows from definition of derivative. $$f_X(x) = F^{'}_X(x) = \lim\limits_{\triangle x \to 0} \frac{F_X(x+\triangle x) - F_X(x)}{\triangle x} = \lim\limits_{\triangle x \to 0} \frac{\mathbb{P}(x \le X \le x+\triangle x)}{\triangle x} .$$ Then $$\lim\limits_{\triangle x \to 0} \left[ \frac{\mathbb{P}(x \le X \le x+\triangle x)}{\triangle x} - f_X(x) \right] = 0.$$ From this we have that $$\frac{\mathbb{P}(x \le X \le x+\triangle x)}{\triangle x} -f_X(x) = o(1),$$ where $o(1)$ is an infinitesimal quantity. Multiplying this equality by $\triangle x$ we get the required.