Answer
(a) True
(b) True
(c) True
Work Step by Step
Since $1/Q\left( {x,y} \right)$ is continuous for all $\left( {x,y} \right)$, by definition we have
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} \frac{1}{{Q\left( {x,y} \right)}} = \frac{1}{{Q\left( {a,b} \right)}}$
for each point $\left( {a,b} \right)$ in its domain.
By Quotient Law of Theorem 1 and the definition of continuity, we have
$\frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} 1}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} Q\left( {x,y} \right)}} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} \frac{1}{{Q\left( {x,y} \right)}} = \frac{1}{{Q\left( {a,b} \right)}}$
$\frac{1}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} Q\left( {x,y} \right)}} = \frac{1}{{Q\left( {a,b} \right)}}$
Thus, $\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} Q\left( {x,y} \right) = Q\left( {a,b} \right)$
Hence, $Q\left( {x,y} \right)$ is continuous for all $\left( {x,y} \right)$. Thus, statement (a) is True.
Since $1/Q\left( {x,y} \right)$ is continuous for all $\left( {x,y} \right)$, it also includes the point $\left( {x,y} \right) \ne \left( {0,0} \right)$. Thus, statement (b) is True.
Since $1/Q\left( {x,y} \right)$ is continuous for all $\left( {x,y} \right)$, $Q\left( {x,y} \right) \ne 0$ for all $\left( {x,y} \right)$. Thus, statement (c) is True.
