# Limit Theorems

1. Feb 4, 2014

### Mathos

1. The problem statement, all variables and given/known data

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

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

we have lim f(xn) = A.

2. Relevant equations

For any ε > 0, |ƒ(x)-A| < ε if x < N

3. 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 xn0 > N if n≥ n0

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

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

2. Feb 5, 2014

### pasmith

The second inequality is backwards.

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

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

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 $R > 0$ there exists $n \in \mathbb{N}$ such that $x_n > R$"). Clearly if $x_n \to \infty$, as we are assuming, then it must follow that $(x_n)$ is not bounded above.

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 $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that for all $n \in \mathbb{N}$, if $n \geq N$ then $|f(x_n) - A| < \epsilon$.

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

3. Feb 5, 2014

### Mathos

Thanks.

Last edited: Feb 5, 2014