Answer
$y=\frac{x+3}{(x+2)(x-5)}$ is continuous as $x\in(-\infty,-2)\cup(-2,5)\cup(5,\infty)$
Work Step by Step
$$y=\frac{x+3}{x^2-3x-10}=\frac{x+3}{(x+2)(x-5)}$$
- Domain: $(-\infty,-2)\cup(-2,5)\cup(5,\infty)$
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 $x=-2$ and $x=5$: $$\lim_{x\to-2}y=\lim_{x\to-2}\frac{x+3}{(x+2)(x-5)}$$ $$\lim_{x\to5}y=\lim_{x\to5}\frac{x+3}{(x+2)(x-5)}$$
As $x\to-2$, $(x+2)\to0$, meaning that $(x+2)(x-5)$ will approach $0$ and $\frac{x+3}{(x+2)(x-5)}$, hence, will approach infinity, not any single value.
Similarly, as $x\to5$, $(x-5)\to0$, meaning that $(x+2)(x-5)$ will approach $0$ and $\frac{x+3}{(x+2)(x-5)}$, hence, will approach infinity again.
In other words, $\lim_{x\to-2}\frac{x+3}{(x+2)(x-5)}$ and $\lim_{x\to5}\frac{x+3}{(x+2)(x-5)}$ do not exist.
So the function is discontinous at $x=-2$ and $x=5$.
Therefore, $y=\frac{x+3}{(x+2)(x-5)}$ is continous in its domain, $(-\infty,-2)\cup(-2,5)\cup(5,\infty)$
