Do Limits Exist for f(x) = 1 at x=5?

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Homework Help Overview

The discussion revolves around the existence of limits for the function f(x) defined as 1 when x is not equal to 5 and 0 when x equals 5. Participants are exploring the implications of continuity and the behavior of limits as x approaches 5.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the limit of f(x) as x approaches 5 and whether it is equal to 1 or not. There are questions about the continuity of the function at x=5 and the relevance of the function's value at that point. Some participants express confusion regarding the application of the epsilon-delta definition of limits.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have offered insights into the nature of limits and continuity, while others are questioning the assumptions made about the function's behavior. There is no explicit consensus yet, but productive dialogue is occurring.

Contextual Notes

Participants are grappling with the definitions of continuity and limits, particularly in the context of piecewise functions. There are references to specific exercises and definitions that may influence their understanding.

Andrax
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This has confused me a bit
I'm typing on my phone so I can't use math symbols sorry
F(x) = 1 ; x=5
0 otherwise (x=/5)
Does lim X approaches 5 f(x) exists and = 1(f continuous at 5) ? Or does it not(because when we approach a were also in the x=/a space so there is also 1)
can someone be really kind and Prove it using epsilon?
I know you can just put alpha = any number you want since 0<epsilon always
I think I'm confusing these functions with the ones like sinx..
 
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Andrax said:
F(x) = 1 ; x=a
0 otherwise (x=/a)
:
when we approach a were also in the x=/a space so there is also 1)
Not at all sure what you mean by that last statement. When x ≠ a, F(x) = 0, yes? What is F(1/n), n = 1, 2, ...?
 
haruspex said:
Not at all sure what you mean by that last statement. When x ≠ 5, F(x) = 0, yes? What is F(1/n), n = 1, 2, ...?
I'm sorry forgot to set the integer fixed
 
F(x) : R--) R
0 x=5
1 x=/5
Is f continuous at 0?
 
Andrax said:
F(x) : R--) R
0 x=5
1 x=/5
Is f continuous at 0?

What is the *definition* for f(x) to be continuous at x = 0? Does your f satisfy that definition? Show your work!
 
The point of this exercise is to make the point that the value of a function at the value of the limit does not matter.
That is

$$\lim _{x \rightarrow a} \mathop{f}(x) $$
Does not depend on the value of f(a)

Perhaps there is a function G that you know of such that
G(x)=A if x=5
G(x)=F(x) otherwise (x=/5)
Choose G(5) whatever you like. So that you know the limit.
 
Ray Vickson said:
What is the *definition* for f(x) to be continuous at x = 0? Does your f satisfy that definition? Show your work!

I'm sorry but I had a huge confusion with this, I'm not sure if when it approaches 5 is it f(5) or not..
I used epsilon delta but. I can't decide what f(x) should I be using
lx-5l<alpha. lf(x)-f(5)l< epsilon..
What f(x) should I use
Just to mention this is not my first time with limits but out of nowhere I found myself unable to do this..
 
lurflurf said:
The point of this exercise is to make the point that the value of a function at the value of the limit does not matter.
That is

$$\lim _{x \rightarrow a} \mathop{f}(x) $$
Does not depend on the value of f(a)

Perhaps there is a function G that you know of such that
G(x)=A if x=5
G(x)=F(x) otherwise (x=/5)
Choose G(5) whatever you like. So that you know the limit.

What are you saying is that the limit of my function x approaches 5 is. 1 and not 0?
We can choose g(5)=100 that doesn't mean limit x approaches 5 is 100?
 
thanks everyone I've got it the limit = 1 => f is not continious at 5
this whole problem was caused by an exercise @ spivak's calculus it was f is discon tinious at l but i thought it is continious .
 

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