Answer
$M(x)$ is continuous on $(-\infty,-1]\cup(0,\infty)$.
Work Step by Step
$$M(x)=\sqrt{1+\frac{1}{x}}$$
Let $f(x)=1+\frac{1}{x}$ and $g(x)=\sqrt x$, we have$$M(x)=g(f(x))$$
We see that $f(x)$ and $g(x)$ are both continuous.
Therefore, by Theorem 9, $M(x)$ is continuous on its domain.
*Find the continuity range of $M(x)$
$M(x)$ is defined when $$1+\frac{1}{x}\ge0$$$$\frac{x+1}{x}\ge0\hspace{.5cm}(x\ne0)$$$$\left\{\begin{array} {c l}x+1\ge0\\x>0\end{array}\right.\hspace{0.5cm}or\hspace{0.5cm}\left\{\begin{array} {c l}x+1\le0\\x<0\end{array}\right.$$$$\left\{\begin{array} {c l}x\ge-1\\x>0\end{array}\right.\hspace{0.5cm}or\hspace{0.5cm}\left\{\begin{array} {c l}x\le-1\\x<0\end{array}\right.$$$$x\gt0\hspace{.5cm}or\hspace{.5cm}x\le-1$$
Therefore, the domain of $g(f(x))$ is $(-\infty,-1]\cup(0,\infty)$.
In conclusion, $M(x)$ is continuous on $(-\infty,-1]\cup(0,\infty)$.
