Continuity of Functions: Proving the Equivalence of Lim f(x) and Lim f(xn)

[Mathos]

Homework Statement

Theorem: Let f:[a, ∞)→ R. The following are equivalent.

i) lim ƒ(x) = A as x→∞
ii) For all sequences {x_n
in [a,∞) with lim x_n = ∞

we have lim f(x_n) = A.

Homework Equations

For any ε > 0, |ƒ(x)-A| < ε if x < N

The Attempt at a Solution

I probably have this wrong, but I think I should show that for any N > 0 in [a, ∞) there exists an x_n0 > N if n≥ n₀

I imagine that to tie this into the idea of continuity, I'd have to come up with an arbitrary function f(c) to get |f(x_n)-f(c)| < ε when x > N

I just don't know how to say f(x) has the same domain as f(x_n) without just stating it.

 

[pasmith]

Homework Statement

Theorem: Let f:[a, ∞)→ R. The following are equivalent.

i) lim ƒ(x) = A as x→∞
ii) For all sequences {x_n
in [a,∞) with lim x_n = ∞

we have lim f(x_n) = A.

Homework Equations

For any ε > 0, |ƒ(x)-A| < ε if x < N

The second inequality is backwards.

f(x) \to A as x \to \infty if and only if for all \epsilon &gt; 0 there exists R &gt; 0 such that for all x \in [a, \infty), if x &gt; R then |f(x) - A| &lt; \epsilon.

x_n \to \infty if and only if for all R &gt; 0 there exists N \in \mathbb{N} such that for all n \in \mathbb{N}, if n \geq N then x_n &gt; R.

The Attempt at a Solution

I probably have this wrong, but I think I should show that for any N > 0 in [a, ∞) there exists an x_n0 > N if n≥ n₀

That, sadly, is wrong. It looks like a conflation of the definitions of "x_n \to \infty" (which I have given above) and "the sequence (x_n) is not bounded above" (which is "for all R &gt; 0 there exists n \in \mathbb{N} such that x_n &gt; R"). Clearly if x_n \to \infty, as we are assuming, then it must follow that (x_n) is not bounded above.

I imagine that to tie this into the idea of continuity, I'd have to come up with an arbitrary function f(c) to get |f(x_n)-f(c)| < ε when x > N

This exercise has nothing to do with continuity. A function is continuous at c if and only if \lim_{x \to c} f(x) = f(c). But this exercise is concerned with the limit as x \to \infty, and since \infty is not a real number there is no such thing as f(\infty). Instead we are given \lim_{x \to \infty} f(x) = A.

To show that (i) implies (ii), you must assume that f(x) \to A as x \to \infty, then take an arbitrary x_n \to \infty and show that for all \epsilon &gt; 0 there exists an N \in \mathbb{N} such that for all n \in \mathbb{N}, if n \geq N then |f(x_n) - A| &lt; \epsilon.

Proving that (ii) implies (i) is slightly harder.

 

[Mathos]

Thanks.