# Limits and Sequences

1. Sep 21, 2013

### Yagoda

1. The problem statement, all variables and given/known data
The function f is defined on a neighborhood N of $\bar{x}$. Show that
$\lim_{x \rightarrow \bar{x}} f(x) = L$ if and only if $\lim_{n \rightarrow \infty} f(x_n) = L$ when $\{x-n\}$ is a sequence of points in N with $\lim_{n \rightarrow \infty} x_n = \bar{x}$.

2. Relevant equations

3. The attempt at a solution
I think I have the necessity, but I am having trouble tightening up the other direction. What I have is that if $\lim_n \rightarrow \infty x_n = \bar{x}$ then $\forall \delta > 0, \exists N_1 : |x_n - \bar{x}| < \delta \text{ when } n > N_1$ and that $\forall \epsilon > 0, \exists N_2 : |f(x_n) - L| < \epsilon \text{ when } n > N_2$.
Let $N = \max\{N_1,N_2\}$ there is an N so that if $|x_n - \bar{x}| < \delta$ then $|f(x_n) - L| < \epsilon$ for any $\epsilon < 0$. Which is close to the definition of the limit of f, but how can I generalize it to $|f(x) - L| < \epsilon$ rather than $|f(x_n) - L| < \epsilon$ or is that not an issue?

2. Sep 21, 2013

### Office_Shredder

Staff Emeritus
Try a proof by contradiction. Suppose that we have for every sequence
$$\lim_{n\to \infty} f(x_n) = L$$,
but that
$$\lim_{x\to \overline{x}} f(x) \neq L$$.

Can you use these two pieces of information to form a contradiction?

3. Sep 21, 2013

### vela

Staff Emeritus
Is this really true? What if the sequence were given by $x_n = \bar{x}$ and f is defined such that f(x)=L if $x=\bar{x}$ and $f(x)=L-1$ for $x\ne\bar{x}$? Clearly, the limit of the sequence f(xn) is L, but the limit of f(x) is L-1. Did you leave out a condition on f, like f being continuous?

4. Sep 22, 2013

### Yagoda

Vela, I checked again and the problem doesn't state that f is continuous, only that it is defined on a deleted neighborhood of $\bar{x}$. Would the limit of the sequence f(xn) really be L in this case though? Intuitively it seems like it would still be L-1 because the x's are arbitrarily close to $\bar{x}$, but not equal to it, but maybe I'm not seeing it correctly.

5. Sep 22, 2013

### Office_Shredder

Staff Emeritus
vela, your function has a limit of L-1 both for the sequences and for the arbitrary $x \to \overline{x}[/tex]. In fact the point of the question really is that it gives you a way of determining that a function is discontinuous. If you know f(x) = L, all you need to do is find a single sequence converging to x such that f(xn) is not equal to L in order to show that f is discontinuous. This will be the most common use of this result that you come across. 6. Sep 22, 2013 ### vela Staff Emeritus How can the limit be L-1 if $f(x_n) = f(\bar{x}) = L$ for all n? Isn't that what I just did with the sequence I suggested? And how are you supposed to use the result to show that f is discontinuous if it supposedly holds for all functions? The OP wasn't asked to prove the statement holds for only continuous functions but for seemingly all functions. 7. Sep 22, 2013 ### christoff The OP likely has some definition of what the expression [itex]\lim_{x\rightarrow\overline x}f(x)=L$ means. This question is merely asking them to show that whatever definition they have is equivalent to having $f(x_n)$ converge to $L$ for any sequence $x_n\rightarrow\overline x$.

We (those of us posting) know that this is equivalent to $f$ being continuous at $\overline x$, but the OP might not have seen this yet.

To answer the OP's original question:
Take Office_Shredder's advice. There is a fairly straightforward proof by contradiction you could try here.

8. Sep 22, 2013

### pasmith

The crucial point is this:

It would have been helpful if the OP had said this initially, but now we know. It follows that your counterexample is not a counterexample, because none of the relevant sequences attain the value $\bar x$, and $f(\bar x)$ is itself undefined.

9. Sep 22, 2013

### vela

Staff Emeritus
Ah, that's what a deleted neighborhood is. I never heard that term before. I thought it was a typo on the OP's part.