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TFAE proof involving limit and convergent sequence


New member
Jun 26, 2019
Let A ⊆ R, let f : A → R, and suppose that (a,∞) ⊆ A for some a ∈ R. Then the
following statements are equivalent:
i) limx→∞ f(x) = L
ii) For every sequence (xn) in A ∩ (a,∞) such that lim(xn) = ∞, the sequence (f(xn))
converges to L.

Not even sure how to begin this one, other than the fact that proving i) -->ii) and ii)--> i) will be sufficient. Could anyone help me with these 2 parts?



Well-known member
MHB Math Helper
Jan 29, 2012
For any proof you start by thinking about what the "hypotheses" and "conclusion" mean: the specific definitions.

For (i)-> (ii) we have the hypothesis [tex]\lim_{x\to \infty} f(x)= L[/tex] so start with the definition of that: Given any [tex]\epsilon> 0[/tex] there exist K such that if x> K then [tex]|f(x)- L[/tex]. And the definition of the conclusion, "for any sequence [tex]\{a_m\}[/tex] in [tex]A\cap (a, \infty)[/tex], such that [tex]\lim a_n=\infty[tex], [tex]\lim f(a_n)= L[/tex]": Given any [tex]\epsilon> 0[/tex] there exist N such that if n> N then [tex]|f(a_n)- L|< \epsilon[/tex].

Now combine the two: Given any [tex]\epsilon> 0[/tex], let [tex]\{a_n\}[/tex] be a sequence in [tex]A\cap (a, \infty)[/tex] such that [tex]\lim a_n= \infty[/tex]. From the hypothesis, there exist K such that if x> K, [tex]|f(x)- L|< \epsilon[/tex]. By the definition of "[tex]\lim a_n= \infty[/tex], there exist N such that if n> N, [tex]a_n> K[/tex]. So there exist N such that if n> N [tex]|f(a_n)- L|< \epsilon[/tex] and we are done. (ii)-> (i) is done similarly.

Important point: definitions in mathematics are "working" definitions- you use the precise wording of definitions in proofs.