# TFAE proof involving limit and convergent sequence

#### brooklysuse

##### New member
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?

Thanks!

#### HallsofIvy

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

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

Now combine the two: Given any $$\epsilon> 0$$, let $$\{a_n\}$$ be a sequence in $$A\cap (a, \infty)$$ such that $$\lim a_n= \infty$$. From the hypothesis, there exist K such that if x> K, $$|f(x)- L|< \epsilon$$. By the definition of "$$\lim a_n= \infty$$, there exist N such that if n> N, $$a_n> K$$. So there exist N such that if n> N $$|f(a_n)- L|< \epsilon$$ 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.