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Limits and Sequences

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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?
 

Answers and Replies

  • #2
Office_Shredder
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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
vela
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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
[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
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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
Office_Shredder
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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
vela
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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?

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
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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
pasmith
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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:

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
vela
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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.
 

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