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

$y=\frac{1}{(x+2)^2}+4$ is continous as $x\in(-\infty,-2)\cup(-2,\infty)$
$$y=\frac{1}{(x+2)^2}+4$$ - Domain: $(-\infty,-2)\cup(-2,\infty)$ We would find all the points where the function is discontinous first, then all the remaining points would be where the function is continous. We see that at $x=-2$: $$\lim_{x\to-2}y=\lim_{x\to-2}\Big(\frac{1}{(x+2)^2}+4\Big)=\lim_{x\to-2}\frac{1}{(x+2)^2}+4$$ As $x\to-2$, $(x+2)^2\to0$, meaning that $\frac{1}{(x+2)^2}$ will approach infinity, not any single value. In other words, $\lim_{x\to-2}\frac{1}{(x+2)^2}$ does not exist. So the function is discontinous at $x=-2$. That is also the only point where $y$ is discontinous. Therefore, $y=\frac{1}{(x+2)^2}+4$ is continous in its domain, $(-\infty,-2)\cup(-2,\infty)$