I have to proof that $\lim_{x \to \c} frac{1}{f(x) = 0$

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Homework Help Overview

The discussion revolves around proving the limit statement \(\lim_{x \to c} \frac{1}{f(x)} = 0\) given that \(f\) is a function mapping a domain \(D\) onto the real numbers, with \(c\) as a limit point where \(\lim_{x \to c} f(x) = \infty\) and \(f(x) \neq 0\) for all \(x\) in \(D\).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of limits and the implications of \(f(x)\) diverging to infinity. There are attempts to relate the limit definition to the inequality involving \(\epsilon\) and \(\delta\). Some participants express uncertainty about how to proceed with the proof.

Discussion Status

Participants are exploring the relationship between the limit of \(f(x)\) as \(x\) approaches \(c\) and the limit of \(\frac{1}{f(x)}\). Some guidance has been offered regarding the definitions needed to support the proof, and there is acknowledgment of the need to clarify the implications of the definitions being used.

Contextual Notes

There is mention of constraints related to the definitions provided in the textbook, particularly regarding continuity and limits at infinity, which may affect how participants approach the proof.

kasperrepsak
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Homework Statement


Given is: f is a function that maps D onto the Real numbers, and c is within D and it is a limit point, and f(x) =/= 0 for all x in D, and [itex]\lim_{x \to c} f(x) = \infty[/itex]
I have to proof that:
[itex]\lim_{x \to c} \frac{1}{f(x)} = 0[/itex]

The Attempt at a Solution



This means that according to the definition I have to proof that [itex]\forall \epsilon \ \exists \delta[/itex] so that [itex]\forall x \in D \ with \ 0 < |x-c|< \delta \ \ : |\frac{1}{f(x)} - 0|< \epsilon[/itex].

Im not sure how to go on from here. Or do i have to do something else?
 
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Hint:
[tex]\left|\frac{1}{f(x)} - 0\right| < \epsilon[/tex]
is equivalent to
[tex]|f(x)| > \frac{1}{\epsilon}[/tex]
 
Would it then be enough to say that because f diverges to infinity as x approaches c, this by definition means that there must be a [itex]\delta[/itex] for which if [itex]0<|x-c|<\delta[/itex] the latter inequality that you wrote holds, which is equivalent to the former, quod erat demonstrandum?
 
Last edited:
kasperrepsak said:

Homework Statement


Given is: f is a function that maps D onto the Real numbers, and c is within D and it is a limit point, and f(x) =/= 0 for all x in D, and [itex]\lim_{x \to c} f(x) = \infty[/itex]
I have to proof that:
[itex]\lim_{x \to c} \frac{1}{f(x)} = 0[/itex]

The Attempt at a Solution

What you have to prove is what you have in the following statement .
This means that according to the definition I have to proof that [itex]\forall \epsilon \ \exists \delta[/itex] so that [itex]\forall x \in D \ with \ 0 < |x-c|< \delta \ \ : |\frac{1}{f(x)} - 0|< \epsilon[/itex].

I'm not sure how to go on from here. Or do i have to do something else?
What will help you get there is to state what is meant by:

[itex]\displaystyle \lim_{x \to c} f(x) = \infty\ .[/itex]
 
Well the weird thing is that in our book we hv only defined what is meant by a sequence diverging to infinity at a point c. This assignment is part of the chapter on continuity where we've only defined what is meant by a limit of f as x goes to infinity. But wut i wrote should b ok since that follows from the definition right?
 
kasperrepsak said:
Well the weird thing is that in our book we have only defined
what is meant by a sequence diverging to infinity at a point c. This assignment is part of the chapter on continuity where we've only defined what is meant by a limit of f as x goes to infinity. But what i wrote should be ok since that follows from the definition right?
What you wrote is what needs to be proved.

What you need, among other things, to get there is a definition of what it means for [itex]\displaystyle \lim_{x \to c} f(x) = \infty\ .[/itex]

You may be able to deduce this from
what is meant by a sequence diverging to infinity at a point c.​
and from
what it means for the limit of a function, f(x), to converge as x → c .​

Basically, what it means for [itex]\displaystyle \lim_{x \to c} f(x) = \infty\,,[/itex] is that given any M > 0, (usually M is a large number) there exists a δ > 0 such that for any x satisfying 0<|x-c|<δ, you have that f(x) > M .
 
No but i ment what i wrote in reply to the firs post. There must be a delta greater then 1/epsilon
 
kasperrepsak said:
No but i meant what i wrote in reply to the firs post. There must be a delta greater then 1/epsilon
That works, if you spell out why it works.

Given an ε > 0, then 1/ε > 0 . Since [itex]\displaystyle \ \lim_{x \to c} f(x) = \infty\ \[/itex] there exists δ > 0, such that ... f(x)>1/ε ...

Added in Edit:

I removed the absolute value from f(x) above
 
Last edited:
OK thank you. Yeah I know that in writing a proof one has to write a lot of little things to make it formally correct, I just wanted to know if I understand the method of proofing. Thanks again : )
 

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