Real Analysis: Proving Lim(yn)=0 from Lim(xn)=Infinity & Lim(xnyn)=L

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Homework Help Overview

The discussion revolves around proving that the limit of the sequence (yn) approaches 0, given that the limit of (xn) approaches infinity and the limit of the product (xnyn) approaches a real number L. The subject area is real analysis, specifically dealing with limits of sequences.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore various approaches to the proof, including examining the implications of the limits provided. Questions arise about the assumptions regarding the sequences and how to construct the proof effectively.

Discussion Status

The discussion is ongoing, with participants offering hints and suggestions for approaching the proof. Some guidance has been provided regarding the limits of related sequences, but there is no explicit consensus on the method to be used.

Contextual Notes

Participants are navigating the complexities of the problem, including the definitions of the sequences and the implications of the limits. There is a noted uncertainty about how to appropriately set up the proof and what assumptions can be made.

rohitmishra
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Question : Let (xn) and (yn) be sequences of real numbers such that lim(xn)= infinity and lim(xnyn)=L for some real number L.

Prove Lim(yn)=0.


I've been trying to solve this question for a long time now. I've no success yet. Can anyone guide me as to how i can approach it.
 
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welcome to pf!

hi rohitmishra! welcome to pf! :smile:

(have an infinity: ∞ and try using the X2 icon just above the Reply box :wink:)

hint: can you prove what limn->∞ 1/xn is? :smile:
 
You know |x_n*y_n - L| < epsilon. Why not see what you can conclude if you let epsilon = |L/2|?
 
I have lim(xnyn) =L.

How can I start of a prove assuming two sequences That is x(n) = 2x(n) which wud tend to infinity and Lim (yn) = 1/x(n) = 0.

Can you tell me what the Hypothesis wud be ?
 
limn->∞ 1/xn = 0
 
ok so you know lim 1/xn and you know lim xnyn, so … ? :smile:
 
Can you help me how i can construct the proof.. ? I understand what you meant. But, How can I assume my yn is 1/xn ?
 
rohitmishra said:
I understand what you meant. But, How can I assume my yn is 1/xn ?

no you don't :redface:, because it isn't

try again :smile:
 

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