## Calculus: Early Transcendentals 8th Edition

Prove that $\lim\limits_{x\to a}f(x)=f(a)$ for $\forall a\in[4,\infty)$.
*NOTES TO REMEMBER: $f(x)$ is continuous on the interval if and only if it is continuous at every point in the interval. In other words, $f(x)$ is continuous on the interval $(u,v)$ if and only if for $\forall a\in(u,v)$, we have $$\lim\limits_{x\to a}f(x)=f(a)$$ For $\forall a\in[4,\infty)$, we consider $\lim\limits_{x\to a}f(x)$ $=\lim\limits_{x\to a}(x+\sqrt{x-4})$ $=\lim\limits_{x\to a}x+\lim\limits_{x\to a}\sqrt{x-4}$ $=\lim\limits_{x\to a}x+\sqrt{\lim\limits_{x\to a}(x-4)}$ $=\lim\limits_{x\to a}x+\sqrt{\lim\limits_{x\to a}x-\lim\limits_{x\to a}4}$ $=a+\sqrt{a-4}$ $=f(a)$ Therefore, $f(x)$ is continuous on the interval $[4,\infty)$.