I need to prove or disprove the following claim: if $$\lim_{x\to x_0} f(x)=L$$ and the set $$A=\lbrace f(x): x\in \Bbb R \rbrace$$ is bounded from above, then $$sup(A)\geq L$$ I tried proving by assuming towards contradiction that $L>sup(A)$ and use the epsilon property of $sup(A)$ but was not able to move on from there. I also tried to disprove the claim with some split functions such that will have $L>f(x)$ for every $f(x) \in A$ but could not find a way to prove the limit. Assistance will be welcomed.
2026-04-09 02:34:38.1775702078
Proving of disproving a claim regarding limits
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Aiming for a contradiction, suppose that $S=\sup(A)<L$. Let $$\varepsilon=\frac{L-S}{2}>0$$ On the one hand, $$f(x)\leq S\quad\text{for all }x\in\mathbb{R}$$ On the other hand, the interval $$(L-\varepsilon,L+\varepsilon) $$ must contain values of $f$ because of the given limit. Can you see the contradiction between the two? (Hint: compare the numbers $L-\varepsilon$ and $S$.)