Limits at infinity, lim xF(x) = L then lim (f(x)=0

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The discussion centers on proving that if \( F:(a,∞) \to \mathbb{R} \) satisfies \( \lim_{x \to \infty} xF(x) = L \) for a finite \( L \), then \( \lim_{x \to \infty} F(x) = 0 \). Participants clarify the proof structure, emphasizing the need for a proper ε-M definition to demonstrate the limit. Key points include establishing that \( |F(x)| < \frac{|L| + 1}{x} \) for sufficiently large \( x \), leading to the conclusion that \( F(x) \) approaches zero as \( x \) approaches infinity.

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Homework Statement



show that if F:(a,∞) -->R is such that lim xF(x) = L, x --> ∞, where L is in R, then lim F(x) = 0, x --> ∞.

Homework Equations


The Attempt at a Solution



Let F:(a,∞) →R is such that lim xF(x) = L, x → infinity, where L is in R. Then there exists an α> 0 where given ε, there exist k(ε) for all x > k then ε > max{1 , ([L]+1)/x} Therefore [xF(x) - L] < 1 whenever x > α. Therefore [F(x)] < ([L]+1)/x. Thus [F(x)-0] < ε Then lim F(x) =0 as x → ∞. This is what I have but it doesn't look right to me.
 
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Proof: let f:(a,∞)→ℝ such that lim xf(x)=L where L in ℝ. Since lim xf(x) = L, there exists α>0 where |xf(x)-L| < 1 for all x > α. Therefore |f(x)|<(|L|+1)/x for x >α. Pick ε = m where there exist δ-neighborhood Vδ(c) of c and x is in A π Vδ(c), there exists m>0, m = |L|+1 then |f(x)| < M, for all X, therefore |f(x)-o|<M=
thus the limitx→∞ f(x) =0.

Does this proof make more sense? Am i still missing something?


kingstrick said:

Homework Statement



show that if F:(a,∞) -->R is such that lim xF(x) = L, x --> ∞, where L is in R, then lim F(x) = 0, x --> ∞.

Homework Equations





The Attempt at a Solution



Let F:(a,∞) →R is such that lim xF(x) = L, x → infinity, where L is in R. Then there exists an α> 0 where given ε, there exist k(ε) for all x > k then ε > max{1 , ([L]+1)/x} Therefore [xF(x) - L] < 1 whenever x > α. Therefore [F(x)] < ([L]+1)/x. Thus [F(x)-0] < ε Then lim F(x) =0 as x → ∞. This is what I have but it doesn't look right to me.
 
kingstrick said:
Proof: let f:(a,∞)→ℝ such that lim xf(x)=L where L in ℝ. Since lim xf(x) = L, there exists α>0 where |xf(x)-L| < 1 for all x > α. Therefore |f(x)|<(|L|+1)/x for x >α. Pick ε = m where there exist δ-neighborhood Vδ(c) of c and x is in A π Vδ(c), there exists m>0, m = |L|+1 then |f(x)| < M, for all X, therefore |f(x)-o|<M=
thus the limitx→∞ f(x) =0.

Does this proof make more sense? Am i still missing something?

You have some undefined quantities.

You can't simply pick the ε in the part where you prove that lim x→∞ f(x) = 0 .

It may help for you to state, in ε - M language, what it means that lim x→∞ f(x) = 0 .
 
This can be proved in one line.

Lt x->infinity
xf(x) =L (where L is finite)

So Lt x->infinity f(x) = L/x (How?)

What do you see??
 
emailanmol said:
This can be proved in one line.

Lt x->infinity
xf(x) =L (where L is finite)

So Lt x->infinity f(x) = L/x (How?)

What do you see??

I think i see L/x going to zero as x goes to infinity since L is finite.
 
Correct :-)
 
SammyS said:
You have some undefined quantities.

You can't simply pick the ε in the part where you prove that lim x→∞ f(x) = 0 .

It may help for you to state, in ε - M language, what it means that lim x→∞ f(x) = 0 .

just curious, what is meant by ε - M language?
 
kingstrick said:
just curious, what is meant by ε - M language?
Given an ε>0, there exists an integer, M, such for all x > M, ...
 
SammyS said:
Given an ε>0, there exists an integer, M, such for all x > M, ...

Thanks... I am an idiot!
 

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