The CDF of a continuous random variable X is: $$ F_X(x) = \begin{cases} 0, & \text{if } x < 4 \\ Ax+B+\frac{C}{x} ,& \text{if } x \geq 4 \end{cases} $$ (a) Find the parameters A, B, C.
Here is how I tried to solve the problem:
$\lim_{x \to \infty}F_x(x) = 1$
{${A = 0, B = 1}$}
Then for the calulation of C my instinct says to just say that $F_X$ is continuous however we have been taught that $F_X$ is only "right" continuous, which does not help me because I can't say: $\lim_{x \to 4^-} = F_X(4)$ and from there find the value of C.
You need $F_X(x)$ to be weakly increasing and, with $\lim\limits_{x \to -\infty}F_X(x) =0$ and $\lim\limits_{x \to +\infty}F_X(x) =1$, never exceeding $1$ or being below $0$.
Your instinct to say that $F_X(x)$ is continuous is wrong in general (it is càdlàg) but correct for a continuous random variable. So here you will need the CDF to be continuous everywhere, in particular at $4$. So indeed $\lim\limits_{x \to 4^-}F_X(x) = F_X(4)$.
Having found $A=0,B=1$, that will make $C=-4$, and so $F_X(x)=1-\frac{4}{x}$ on $x \ge 4$. This will be weakly increasing and have the correct limit of $1$ on the right.
You can go further and say that the density is $$ f_X(x) = \begin{cases} 0, & \text{if } x < 4 \\ \frac{4}{x^2} ,& \text{if } x \geq 4 \end{cases} $$
and that you have a Pareto distribution with shape parameter $\alpha=2$ and minimum $4$.