$f:[2;\infty)\backslash\{3\}\to\mathbb{R}$, $$f(x):=\frac{|x^2-4x+3|}{x^2-2x-3}=\begin{cases}\frac{(x^2-4x+3)}{x^2-2x-3} & x\geq3\\\frac{-(x^2-4x+3)}{x^2-2x-3} & x<3\end{cases}$$
I see $f$ is undefined in $3$ because the denominator has a zero in this point. So my first thought is actually to factorize $f$ by finding the zeros (e.g. with the p-q-formula): $$f(x)=\frac{|x^2-4x+3|}{x^2-2x-3}=\frac{|(x-1)(x-3)|}{(x-3)(x+1)}=\begin{cases}\frac{(x-1)}{(x+1)} & (x-1)(x-3)\geq0\\\frac{-(x-1)}{(x+1)} & (x-1)(x-3)<0\end{cases}$$ Since $1\notin [2;\infty)$, I only need to consider the critical point $x_0=3$.
Now I want to identify when the absolute value function changes its sign. Approaching $3$ from the left gives (since $x<3$ and $x$ close to $3$): $(x-1)(x-3)<0$. So we are in the second case for the left hand limit. For the right hand limit (since $x>3$) we are in the first case , correct? Is this an efficient method to dissolve absolute value functions like this when checking for continuity?
General question: we also factorized functions like $\frac{1}{4-x^2}$ to examine whether the limit is $+\infty$ or $-\infty$ when approaching $x_0=2$ and we factorized functions like $\frac{x^2+5x-24}{x+8}$ just to simplify. So when do I factorize a function when checking for continuity? When the critical value is a zero of the denominator, to simplify or to decide when the absolute value function changes its sign?