# Question about continuous functions and limits

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 $$0, and $$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..

• I presume you'll have $g(1) = 0$, which shouldn't give any problems. – Elchanan Solomon Dec 30 '20 at 10:32
• that's why I first assume toward contradiction g(1) is not 0 – MathCurious Dec 30 '20 at 10:37
• "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..": there is no difference indeed! – Crostul Dec 30 '20 at 11:12
• so can I really justfy using the inequality given to us since my g(1) acctualy uses 1 and not x's that approaching to 1? – MathCurious Dec 30 '20 at 11:38

## 1 Answer

$$g$$ 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 $$g(1) = 0$$.