Limits & Sequences Homework: Show $\lim_{x \rightarrow \bar{x}} f(x) = L$

In summary, the statement states that for a function f defined on a neighborhood N of \bar{x}, the limit of f as x approaches \bar{x} is equal to L if and only if the limit of the sequence f(x_n) is equal to L for any sequence of points x_n in N that converge to \bar{x}. This can be proven by contradiction by assuming that the limit of f as x approaches \bar{x} is not equal to L, and showing that this contradicts the given condition for the sequence f(x_n).
  • #1
Yagoda
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Homework Statement


The function f is defined on a neighborhood N of [itex]\bar{x}[/itex]. Show that
[itex] \lim_{x \rightarrow \bar{x}} f(x) = L [/itex] if and only if [itex] \lim_{n \rightarrow \infty} f(x_n) = L [/itex] when [itex] \{x-n\}[/itex] is a sequence of points in N with [itex]\lim_{n \rightarrow \infty} x_n = \bar{x} [/itex].


Homework Equations





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 [itex]\lim_n \rightarrow \infty x_n = \bar{x} [/itex] then [itex]\forall \delta > 0, \exists N_1 : |x_n - \bar{x}| < \delta \text{ when } n > N_1[/itex] and that [itex]\forall \epsilon > 0, \exists N_2 : |f(x_n) - L| < \epsilon \text{ when } n > N_2 [/itex].
Let [itex] N = \max\{N_1,N_2\} [/itex] there is an N so that if [itex]|x_n - \bar{x}| < \delta [/itex] then [itex]|f(x_n) - L| < \epsilon[/itex] for any [itex]\epsilon < 0[/itex]. Which is close to the definition of the limit of f, but how can I generalize it to [itex]|f(x) - L| < \epsilon[/itex] rather than [itex]|f(x_n) - L| < \epsilon[/itex] or is that not an issue?
 
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  • #2
Try a proof by contradiction. Suppose that we have for every sequence
[tex] \lim_{n\to \infty} f(x_n) = L [/tex],
but that
[tex] \lim_{x\to \overline{x}} f(x) \neq L [/tex].

Can you use these two pieces of information to form a contradiction?
 
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  • #3
Yagoda said:

Homework Statement


The function f is defined on a neighborhood N of [itex]\bar{x}[/itex]. Show that
[itex] \lim_{x \rightarrow \bar{x}} f(x) = L [/itex] if and only if [itex] \lim_{n \rightarrow \infty} f(x_n) = L [/itex] when [itex] \{x_n\}[/itex] is a sequence of points in N with [itex]\lim_{n \rightarrow \infty} x_n = \bar{x} [/itex].

Homework Equations


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 [itex]\lim_{n \rightarrow \infty} x_n = \bar{x} [/itex] then
[tex]\forall \delta > 0, \exists N_1 : |x_n - \bar{x}| < \delta \text{ when } n > N_1[/tex] and that [tex]\forall \epsilon > 0, \exists N_2 : |f(x_n) - L| < \epsilon \text{ when } n > N_2.[/tex] Let [itex] N = \max\{N_1,N_2\} [/itex]. There is an N so that if [itex]|x_n - \bar{x}| < \delta [/itex] then [itex]|f(x_n) - L| < \epsilon[/itex] for any [itex]\epsilon > 0[/itex], which is close to the definition of the limit of f, but how can I generalize it to [itex]|f(x) - L| < \epsilon[/itex] rather than [itex]|f(x_n) - L| < \epsilon[/itex], or is that not an issue?
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
Vela, I checked again and the problem doesn't state that f is continuous, only that it is defined on a deleted neighborhood of [itex]\bar{x}[/itex]. 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 [itex]\bar{x}[/itex], but not equal to it, but maybe I'm not seeing it correctly.
 
  • #5
vela, your function has a limit of L-1 both for the sequences and for the arbitrary [itex] 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
Yagoda said:
Vela, I checked again and the problem doesn't state that f is continuous, only that it is defined on a deleted neighborhood of [itex]\bar{x}[/itex]. 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 [itex]\bar{x}[/itex], but not equal to it, but maybe I'm not seeing it correctly.
How can the limit be L-1 if ##f(x_n) = f(\bar{x}) = L## for all n?

Office_Shredder said:
vela, your function has a limit of L-1 both for the sequences and for the arbitrary [itex] 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.
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
vela said:
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.

The OP likely has some definition of what the expression [itex]\lim_{x\rightarrow\overline x}f(x)=L[/itex] means. This question is merely asking them to show that whatever definition they have is equivalent to having [itex]f(x_n)[/itex] converge to [itex]L[/itex] for any sequence [itex]x_n\rightarrow\overline x[/itex].

We (those of us posting) know that this is equivalent to [itex]f[/itex] being continuous at [itex]\overline x[/itex], 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
vela said:
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.

The crucial point is this:

Yagoda said:
Vela, I checked again and the problem doesn't state that f is continuous, only that it is defined on a deleted neighborhood of [itex]\bar{x}[/itex].

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 [itex]\bar x[/itex], and [itex]f(\bar x)[/itex] is itself undefined.
 
  • #9
pasmith said:
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 [itex]\bar x[/itex], and [itex]f(\bar x)[/itex] is itself undefined.
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.
 

1. What does the notation $\lim_{x \rightarrow \bar{x}} f(x) = L$ mean?

The notation represents the limit of a function f(x) as the input x approaches a specific value $\bar{x}$. The limit is equal to L if the function approaches the same value L from both the left and right sides of $\bar{x}$. This concept is used to describe the behavior of a function near a specific point.

2. How do you calculate limits algebraically?

To calculate a limit algebraically, you can use various techniques such as factoring, rationalizing, and manipulating the expression. The goal is to simplify the expression so that you can either directly substitute the limit value or use basic limit rules to evaluate the limit.

3. Can a function have a limit at a point where it is not defined?

Yes, a function can have a limit at a point where it is not defined. This means that the function is approaching a specific value as the input gets closer and closer to the point, even though the function is not defined at that point.

4. What is the difference between a limit and a sequence?

A limit refers to the behavior of a function near a specific point, while a sequence is a list of numbers that follow a specific pattern or rule. A limit can be defined at any point on a function, while a sequence is a list of values that can be infinite or finite.

5. How do you prove that a limit exists?

To prove that a limit exists, you can use the epsilon-delta definition, which states that for any positive value $\epsilon$, there exists a positive value $\delta$ such that if the distance between the input x and the limit point $\bar{x}$ is less than $\delta$, then the distance between the output f(x) and the limit value L is less than $\epsilon$. This means that as x gets closer to $\bar{x}$, f(x) gets closer to L.

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