## University Calculus: Early Transcendentals (3rd Edition)

The extended function's value at $x=0$ is $7.4$.
$$f(x)=(1+2x)^{1/x}$$ The graph of the function is shown below. For $f$ to have a continuous extension to the origin, $\lim_{x\to0}f(x)$ must exist first. Looking at the graph, $f(x)$ both approaches a value of $7.4$ as $x\to0^+$ and $x\to0^-$. That means $\lim_{x\to0}f(x)$ does exist and equals $7.4$. Therefore, even though $f(0)$ is not defined currently, $f$ has a continuous extension to the origin and can extend to include a value of $x=0$ so that $f(0)=\lim_{x\to0}f(x)$ and then $f$ would be continuous at $x=0$. That being said, the extended function's value at $x=0$ here should be $7.4$.