Limit of a function at an accumulation point

[monkey372]

Homework Statement

Let f : D → R and c ∈ R an accumulation point of D.

Homework Equations

Prove the following are equivalent:[/b]

(a) f has a limit at c.

(b) For all sequences (sn ) such that c = sn ∈ D for all n ∈ N and sn → c,
the sequence (f (sn )) is convergent in R.

The Attempt at a Solution

Here is my approach:
Let x be an arbitrary element of D. Since f has a limit at c which means f(x) -> L as x -> c, we have the sequence xn -> c, where (xn) does not equal to c for all natural number n, as f(xn)->c.

But then I don't know how I should continue. Please help me out.

 

[mighty2000]

Hello! Your approach is a good start. Here's how you can continue:

Since f has a limit at c, we know that for any ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε.

Now, for any sequence (sn) such that c = sn ∈ D for all n ∈ N and sn → c, we know that there exists a positive integer N such that for all n > N, 0 < |sn - c| < δ. This means that for all n > N, |f(sn) - L| < ε.

Since this is true for any ε > 0, we can say that the sequence (f(sn)) converges to L. Therefore, (a) implies (b).

To show that (b) implies (a), we can use a proof by contradiction. Assume that (a) is false, i.e. f does not have a limit at c. This means that there exists an ε > 0 such that for any δ > 0, there exists a point x in D such that 0 < |x - c| < δ but |f(x) - L| ≥ ε, where L is the supposed limit of f at c.

Now, consider the sequence (xn) defined as follows: xn = c + (1/n). Note that xn ∈ D for all n ∈ N and xn → c. However, for this sequence, we have |f(xn) - L| ≥ ε for all n ∈ N, which contradicts the fact that (f(xn)) converges to L. Therefore, our assumption that (a) is false must be incorrect, and hence (b) implies (a).

Hence, we have shown that (a) and (b) are equivalent.