Is $\frac{1}{x}$ strictly positive on $x \in (0, +\infty)$ Consider the function $\frac{1}{x}$ defined on the interval $x \in (0, +\infty)$. Is this function strictly positive $f(x) > 0$? It should be obvious but the fact that $\lim\limits_{x\uparrow+\infty}f(x)=0$ makes me uncomfortable. Is it really allowed to say that the function is strictly positive when there exists a limit which is zero?
 A: As has been pointed out by @MartinR "strictly positive" is not the same as "bounded below by a positive number". Writing this formally: $\forall{f}$ and $\forall{x} \in \Omega$ and $\forall{a} > 0$ the expression $\neg (f(x) > 0 \iff \inf\{f(\Omega)\} = a)$ is true.
To prove by contradiction, assume that there exist such $f$, $x \in \Omega$ and $a > 0$ such that expression $f(x) > 0 \iff \inf\{f(\Omega)\} = a$ leads to a contradiction. Consider forward implication,  $\inf\{f(\Omega)\} = a \iff$ ($\forall{x} \in \Omega$, the inequality $a \leq f(x)$ holds). Choose $f$ and $x$ so that $f(x) = \frac{a}{2} > 0$, then $a \leq \frac{a}{2}$ is a contradiction which concludes the proof.
Considering this, $\frac{1}{x}$ on $(0, +\infty)$ is strictly positive and bounded below by $0$.
A: The key point is that a function is only well-defined if it is provided with a domain and codomain, and its properties depend on its behavior strictly on its domain. See (MSE Q1087412) for an overview of what constitutes a function.
Does the function $f:x\mapsto 2x$ take strictly integer values, strictly rational values, all real values, or all complex values? It depends on its domain. Choosing a domain of $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$ could make any of these true. In fact, you could quite convincingly argue that none of these are even the same function, since the domain is a necessary component of the definition.
And the functions $\sin$ and $\cos$ surely cannot be strictly positive? But they might well be if their domains are both limited to $(0,\pi/2)$.
For a slightly more involved example, we may also ask whether functions that are discontinuous on one domain (say, $\mathbb{R}$) might, counterintuitively, be continuous on another (say, $\mathbb{Q}$). And indeed, they can be, for example $\displaystyle g:x\mapsto \begin{cases}1,&x\ge\sqrt{2}\\0,&x<\sqrt{2}\end{cases}$ or the Dirichlet function $\displaystyle h:x\mapsto \begin{cases}1,&x\,\text{rational}\\0,&\,\text{else}\end{cases}$. See (MSE Q1812344) and (MSE Q228233).
With this in mind, the function $i:x\mapsto\frac1x$ is strictly positive over the domain $(0,+\infty)$ despite not being so over $\mathbb{R}/\{0\}$ or, if we permit the use of the extended reals and define $\frac1{+\infty}=0$, the domain $(0,+\infty]$. It doesn't matter how the function behaves outside of its domain.
