A function satisfying lim h->0[f(x+h-f(x-h)] = 0, and limit at 0 dne

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Homework Help Overview

The discussion revolves around finding or demonstrating the impossibility of a function \( f: \mathbb{R} \to \mathbb{R} \) that satisfies the condition \( \lim_{h \to 0} [f(x+h) - f(x-h)] = 0 \) while having a limit that does not exist at zero.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the properties of piecewise defined functions and question the implications of the limit conditions. There is a discussion about the continuity of the function at points other than zero and how that affects the limit as \( h \) approaches zero.

Discussion Status

Some participants have proposed specific functions, such as \( f(x) = \cos(1/x) \) for \( x \neq 0 \) and \( f(0) = 0 \), and are evaluating whether these meet the criteria set forth in the original problem. There is ongoing exploration of the behavior of these functions and their limits.

Contextual Notes

Participants note the requirement for the function to be defined on \( (0, \infty) \) and the challenge of extending it to \( (-\infty, 0) \) while maintaining the specified limit behavior.

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Homework Statement


Give an example or show why it is impossible
A function f: R→R satisfying lim h->0[f(x+h-f(x-h)] = 0, and limit at 0 dne

Homework Equations



n/a

The Attempt at a Solution



Been trying different piecewise defined functions, with none of them working. I do not really know where to go. I can sit here for hours trying different functions, but i decided to see if anyone on here could help me out.
 
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[edit] oops, misread the question. will update in a minute.
 
DotKite said:

Homework Statement


Give an example or show why it is impossible
A function f: R→R satisfying lim h->0[f(x+h-f(x-h)] = 0, and limit at 0 dne

Homework Equations



n/a

The Attempt at a Solution



Been trying different piecewise defined functions, with none of them working. I do not really know where to go. I can sit here for hours trying different functions, but i decided to see if anyone on here could help me out.

So, in readable English are you saying that ##\lim_{h \to 0} f(x+h)## does not exist?
 
Well, you know you need a function whose limit at zero does not exist. So start with a function which is only defined on ##(0, \infty)## and for which the limit does not exist as ##h \rightarrow 0##. Then consider how you can extend this function to ##(-\infty, 0)## in such a way that the bad behavior will cancel out when you form the difference ##f(x+h) - f(x-h)## at ##x = 0##.
 
Would this work,

f(x) = cos(1/x) x ≠ 0
f(x) = 0 , x = 0
 
DotKite said:
Would this work,

f(x) = cos(1/x) x ≠ 0
f(x) = 0 , x = 0
Yes, that works. ##f(0+h) - f(0-h) = 0## for all positive ##h## but ##\lim_{x \rightarrow 0} f(x)## does not exist. You should also explain why ##\lim_{h \rightarrow 0} [f(x+h) - f(x-h)] = 0## for ##x \neq 0##.
 
jbunniii said:
Yes, that works. ##f(0+h) - f(0-h) = 0## for all positive ##h## but ##\lim_{x \rightarrow 0} f(x)## does not exist. You should also explain why ##\lim_{h \rightarrow 0} [f(x+h) - f(x-h)] = 0## for ##x \neq 0##.

lim h→0 [f(x+h) - f(x-h)] = 0 because as h→0 we get lim f(x+0) - f(x-0) = lim f(x) - f(x) = 0

right?
 
DotKite said:
lim h→0 [f(x+h) - f(x-h)] = 0 because as h→0 we get lim f(x+0) - f(x-0) = lim f(x) - f(x) = 0

right?
That's true as long as ##f## is continuous at ##x##. So the key is that your function is continuous at all ##x \neq 0##.
 

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