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Solution (a) The domain of the exponential function is the set of all real numbers, so $f$ is defined for any number $c$. That is, $f(c) = e^{2c}$. Also for any number $c$, $$\begin{equation*} \lim\limits_{x\rightarrow c}f(x) = \lim\limits_{x\rightarrow c}e^{2x} = \lim\limits_{x\rightarrow c}(e^{x}) ^{2} = \left[ \lim\limits_{\kern.5ptx\rightarrow c}e^{x}\right] ^{2} = (e^{c}) ^{2} = e^{2c} = f(c) \end{equation*}$$

Since $\lim\limits_{x\rightarrow c}f(x) = f(c)$ for any number $c,$ then $f$ is continuous at all numbers $c$.

(b) The logarithmic function $f(x) = \ln x$ is continuous on its domain, the set of all positive real numbers. The function $g(x) = \sqrt[3]{x}$ is continuous on its domain, the set of all real numbers. Then for any real number $c>0$, the composite function $F(x) = (g\circ f) (x) =$ $\sqrt[3]{\ln x}$ is continuous at $c$. That is, $F$ is continuous at all numbers $x>0$.