Answer
$y=\frac{x+1}{x^2-4x+3}$ is continous as $x$ $\in(-\infty,1)\cup(1,3)\cup(3,\infty)$
Work Step by Step
$$y=\frac{x+1}{x^2-4x+3}=\frac{x+1}{(x-1)(x-3)}$$
- Domain: $(-\infty,1)\cup(1,3)\cup(3,\infty)$
We would find all the points where the function is discontinous first, then all the remaining points would be where the function is continous.
- At $x=1$ and $x=3$: $$\lim_{x\to1}y=\lim_{x\to1}\frac{x+1}{(x-1)(x-3)}$$ $$\lim_{x\to3}y=\lim_{x\to3}\frac{x+1}{(x-1)(x-3)}$$
As $x\to1$, $(x-1)\to0$, meaning that $(x-1)(x-3)$ will approach $0$ and $\frac{x+1}{(x-1)(x-3)}$, hence, will approach infinity, not any single value.
Similarly, as $x\to3$, $(x-3)\to0$, meaning that $(x-1)(x-3)$ will approach $0$ and $\frac{x+1}{(x-1)(x-3)}$, hence, will approach infinity again.
In other words, $\lim_{x\to1}\frac{x+1}{(x-1)(x-3)}$ and $\lim_{x\to3}\frac{x+1}{(x-1)(x-3)}$ do not exist.
So the function is discontinous at $x=1$ and $x=3$.
Therefore, $y=\frac{x+1}{x^2-4x+3}$ is continous in its domain, $(-\infty,1)\cup(1,3)\cup(3,\infty)$
