# Real Analysis Proof (Limits of Functions)

1. Jun 24, 2013

### jmjlt88

Let A be a subset of ℝ. Let c be a limit point of A. Consider the function f: A → ℝ

Claim: If the function f has not have a limit at c, then there exists a sequence (xn), where xn≠c for all n, such that lim xn=c, but the sequence (f(xn)) does not converge.

Since the function f does not have a limit at c, for all L in ℝ, there is some εL>0 such that for all δ>0 there is a point x in A satisfying 0<|x-c|<δ and yet |f(x)-L|≥εL.
I know how to create a sequence that converges to 0; I am having trouble making the sequence (f(xn)) not converge.

My initial idea was to pick points xn such that 0<|xn-c|<1/n and |f(xn)-n|≥εn for all n. However, this was not quite working out. Any bump in the right direction would be awesome!

2. Jun 24, 2013

### verty

Hint: the value f(c), if defined, is irrelevant.

I'm trying to hint without being obvious.

3. Jun 24, 2013

### pasmith

Is f assumed to be continuous except at c?

In what ways can a continuous function on an open interval fail to have a limit at an endpoint?

4. Jun 24, 2013

### jmjlt88

To pasmith, no, f is not assumed to be continuous.

To verty, indeed. A couple ways the sequence (f(xn)) could fail to converge is if we have some asymptotic behavior or something like the signum function. Unfortunately, the hint provided is not getting the wheels in motion. :(

5. Jun 24, 2013

### verty

Think about the contrapositive of the claim, it may be easier to prove.

6. Jun 24, 2013

### jmjlt88

The contrapositive of the claim would be:
"If for every sequence (xn) that converges to c, the sequence (f(xn)) converges, then f has a limit at c."

I tried the contrapositive (and also had some difficulty). However, I gave up rather quickly since I am so focus on constructing that sequence to prove the original claim. I would really like to move on (as I am only reviewing the major topics from introductory analysis). As such, another hint concerning the original claim would be fantastic!

7. Jun 25, 2013

### pasmith

How far did you get?

Did you show that the limit of the sequence $(f(x_n))$ is independent of the sequence $(x_n)$?