## University Calculus: Early Transcendentals (3rd Edition)

$y=\frac{x+1}{x^2-4x+3}$ is continous as $x$ $\in(-\infty,1)\cup(1,3)\cup(3,\infty)$
$$y=\frac{x+1}{x^2-4x+3}=\frac{x+1}{(x-1)(x-3)}$$ - 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, $(-\infty,1)\cup(1,3)\cup(3,\infty)$