TFAE proof involving limit and convergent sequence

[brooklysuse]

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]

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 [tex]\lim a_n=\infty[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.

 

[soloenergy]

Sure, here's how you can approach this proof:

i) --> ii):

Assume that limx→∞ f(x) = L. This means that for any ε > 0, there exists a number M such that for all x > M, |f(x) - L| < ε. Now, consider a sequence (xn) in A ∩ (a,∞) such that lim(xn) = ∞. This means that for any M, there exists an N such that for all n > N, xn > M. Since (a,∞) ⊆ A, this also means that for all n > N, xn ∈ A. Therefore, for all n > N, we have xn > M and xn ∈ A, so xn ∈ A ∩ (a,∞). Now, using the definition of limit, we can say that for any ε > 0, there exists an N such that for all n > N, |xn - ∞| < ε. Since xn > M for all n > N, this means that for all n > N, xn > M and |f(xn) - L| < ε. Therefore, by the definition of convergence, we can say that lim(f(xn)) = L.

ii) --> i):

Assume that for every sequence (xn) in A ∩ (a,∞) such that lim(xn) = ∞, the sequence (f(xn)) converges to L. This means that for any ε > 0, there exists an N such that for all n > N, |f(xn) - L| < ε. Now, consider any x > a. Since (a,∞) ⊆ A, this means that x ∈ A. Now, consider a sequence (xn) in A ∩ (a,∞) such that xn = x + n. This sequence satisfies the condition that lim(xn) = ∞. Therefore, by our assumption, we can say that lim(f(xn)) = L. But since xn = x + n, this means that for all n > N, |f(x + n) - L| < ε. Now, using the definition of limit, we can say that limx→∞ f(x) = L.

Therefore, we have shown that i) and ii) are equivalent.