Show that a Function Is Continuous

Show that:

1. \(f(x) = e^{2x}\) is continuous for all real numbers.
2. \(F(x) = \sqrt[3]{\ln x}\) is continuous for \(x>0\).

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 \(x>0\).