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

[rohitmishra]

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.

 

[tiny-tim]

welcome to pf!

hi rohitmishra! welcome to pf!

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

hint: can you prove what lim_n->∞ 1/x_n is?

 

[platinumtucan]

You know |x_n*y_n - L| < epsilon. Why not see what you can conclude if you let epsilon = |L/2|?

 

[rohitmishra]

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 ?

 

[rohitmishra]

limn->∞ 1/xn = 0

 

[tiny-tim]

ok so you know lim 1/x_n and you know lim x_ny_n, so … ?

 

[rohitmishra]

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 ?

 

[tiny-tim]

I understand what you meant. But, How can I assume my yn is 1/xn ?

no you don't , because it isn't

try again ​