Answer
$$D_f= [-c^2, \infty ) - \{ 0 \}$$
$$f(c)=\frac{1}{2c}$$
Work Step by Step
To find the domain of $f$, we should consider the following points:
First, we must exclude points vanishing the denominator, so $x \neq 0$; second, we must exclude points making the radicand negative, so $x+c^2 \ge 0 \quad \Rightarrow \quad x \ge -c^2$. Thus, the domain of $f$ is$$D_f=[-c^2 , \infty ) - \{0 \}.$$To make $f$ continuous at $x=0$, we must have $f(0)=\lim_{ x \to 0 }f(x)$, so we first need to calculate $\lim_{ x \to 0 }f(x)$.
Since we get the indeterminate form $\frac{0}{0}$ by direct substitution, we have to rationalize the numerator:$$\frac{\sqrt{x+c^2}-c}{x}=\left (\frac{\sqrt{x+c^2}-c}{x} \right ) \left ( \frac{\sqrt{x+c^2}+c}{\sqrt{x+c^2}+c} \right )=\frac{x}{x(\sqrt{x+c^2}+c)}=\frac{1}{\sqrt{x+c^2}+c}$$(Please note that $x \neq 0$). Hence,$$\lim_{ x \to 0} \frac{\sqrt{x+c^2}-c}{x}=\lim_{x \to 0} \frac{1}{\sqrt{x+c^2}+c}=\frac{1}{2c}$$(Please note that in the last equality we have used the fact $c >0$).
Thus, to make $f$ continuous at $x=0$, we must define $f$ at $x=c$ as $f(c)=\frac{1}{2c}$.
