The following intuitive example is sometimes given when density functions are introduced (see equation (1) here, for example)
Suppose that $X$ is a random variable with an absolutely continuous distribution and a density function $f$. Then $$ P(x<X\le x+\varepsilon)\approx f(x)\cdot\varepsilon $$ for small values of $\varepsilon>0$ and $x\in\mathbb R$.
I think I intuitively understand the idea behind this example, but is it possible to make this example rigorous? For instance, is it true that $$ \frac{P(x<X\le x+\varepsilon)}{f(x)\cdot\varepsilon}\to1 $$ as $\varepsilon\to0$ provided that $f(x)>0$? Is it possible to establish a similar statement (maybe under additional assumptions)?
Any help is much appreciated!
The question has actually little to do with probabilites.
If you take for granted that the random variable has a $\text{cdf}$, this means that
$$\mathbb P(X\le x)=\text{cdf}_X(x).$$ From there, by the mean value theorem,
$$\mathbb P(x<X\le x+\epsilon)=\text{cdf}_X(x+\epsilon)-\text{cdf}_X(x)= \epsilon\frac d{dx}\text{cdf}_X{(x+\mu\epsilon)}$$ if the $\text{cdf}$ is a differentiable function.
Also
$$\lim_{\epsilon\to0}\frac{\text{cdf}_X(x+\epsilon)-\text{cdf}_X(x)}\epsilon=\frac d{dx}\text{cdf}_X{(x)}=:\text{pdf}_X(x),$$ which you called $f$.