Is continuous function finite-valued in $R^n$? Let $f$ be a continuous function defined on $\mathbb{R}^n$. The range of $f$ is in the extended real numbers. Is $f<\infty \ \forall x\in \mathbb{R}^n$ ? And why?

In the "usual" topology on the extended real numbers $[-\infty, \infty]$, a neighborhood of $\infty$ is a set containing some interval $(a, \infty] = (a, \infty) \cup \{\infty\}$ with $a$ real, and similarly for neighborhoods of $-\infty$.
If that's true in your setting, $[-\infty, \infty]$ is homeomorphic to a closed, bounded interval of real numbers. For instance, the hyperbolic tangent function $\tanh$ is a homeomorphism from $[-\infty, \infty]$ to $[-1, 1]$. Consequently, asking whether a continuous function can achieve the value $\infty$ is no more mysterious than asking whether a continuous, real-valued function can achieve an absolute maximum.
As MPW and Ethan Bolker note, the constant function with value $\infty$ is continuous. Non-constant continuous functions can achieve the values $\infty$ and/or $-\infty$, as well. For example, if $n = 1$, then


*

*$f(x) = \frac{1}{x^{2}}$ (extended by $f(0) = \infty$) is continuous, since $\lim\limits_{x \to 0} \frac{1}{x^{2}} = \infty$.

*$f(x) = \frac{1}{x}$ (extended by $f(0) = \infty$) is not continuous, since $\lim\limits_{x \to 0^{\pm}} \frac{1}{x} = \pm\infty$ (i.e., the one-sided limits exist as extended real numbers, but are not equal).
And so forth.
