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

by kingstrick
Tags: f(x), infinity, limit, x -->
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kingstrick
#1
Mar14-12, 01:06 PM
P: 108
1. The problem statement, all variables and given/known data

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 --> ∞.

2. Relevant equations



3. 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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kingstrick
#2
Mar15-12, 06:16 PM
P: 108
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?


Quote Quote by kingstrick View Post
1. The problem statement, all variables and given/known data

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 --> ∞.

2. Relevant equations



3. 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.
SammyS
#3
Mar15-12, 06:40 PM
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Quote Quote by kingstrick View Post
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 .

emailanmol
#4
Mar15-12, 06:47 PM
P: 297
Limits at infinity, lim xF(x) = L then lim (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??
kingstrick
#5
Mar15-12, 06:53 PM
P: 108
Quote Quote by emailanmol View Post
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.
emailanmol
#6
Mar15-12, 07:17 PM
P: 297
Correct :-)
kingstrick
#7
Mar15-12, 10:04 PM
P: 108
Quote Quote by SammyS View Post
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?
SammyS
#8
Mar16-12, 02:01 AM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
Quote Quote by kingstrick View Post
just curious, what is meant by ε - M language?
Given an ε>0, there exists an integer, M, such for all x > M, ...
kingstrick
#9
Mar16-12, 06:01 AM
P: 108
Quote Quote by SammyS View Post
Given an ε>0, there exists an integer, M, such for all x > M, ...
Thanks... I am an idiot!


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