Question about continuous functions and limits

Question

I posted a thread about a certain problem I was working on and this gap in my knowledge came up, consider for example the function $g: (0,\infty)\to \mathbb R $, for which $g\left(x\right)\cdot g\left(\frac{1}{x}\right)<0$ for every $0<x≠1$, and $g(x)$ is continuous at $1$.

So the way we define continuity is

".. for every $ϵ > 0$ there exists a $δ > 0$ such that for every $|x − a| < δ$, we have $| f(x) − f(a)| < ϵ$."

Therefore, when I calculate the limit of the expression above when x tend to 1, and reach that $g(x)^2$≥0, how can I justify using the given inequality above? because x CAN be 1, and if f(1) is actualy 1, not not tends to 1 then I can't use the inequality? Is the way I'm looking at it not correct? I also thought of a possibility, that since I only calculate the limit and not the actual expression, I would get a positive limit for function that can't be positive! so if I assume f(1) is not 0, I would reach a contradiction.. But still I can't quiet get my head around the limits of continuous functions and the difference between the value of the limit and the expression at the point x tends to..

Answer 1

$g$ is continuous at $1$. By the continuity of $x\mapsto \frac 1x$, $g\left(\frac 1x\right)$ is also continuous at $1$. And by the continuity of multiplication, so is their product $g(x)g\left(\frac 1x\right)$. So $\lim_{x \to 1} g(x)g\left(\frac 1x\right) = g^2(1)$.

But if a function takes its values in a closed set, any convergent limit of the function must also be in that closed set. $(-\infty, 0]$ is closed, so by your inequality, $g^2(1) \le 0$. As the square of a real number $g^2(1) \ge 0$. So the only possibility is $g(1) = 0$.