1
$\begingroup$

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..

$\endgroup$
4
  • $\begingroup$ I presume you'll have $g(1) = 0$, which shouldn't give any problems. $\endgroup$ – Elchanan Solomon Dec 30 '20 at 10:32
  • $\begingroup$ that's why I first assume toward contradiction g(1) is not 0 $\endgroup$ – MathCurious Dec 30 '20 at 10:37
  • $\begingroup$ "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! $\endgroup$ – Crostul Dec 30 '20 at 11:12
  • $\begingroup$ 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? $\endgroup$ – MathCurious Dec 30 '20 at 11:38
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.