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

In summary, if F:(a,∞) →R such that lim xF(x) = L, x → ∞, where L is in R, then lim F(x) = 0, x → ∞. This can be proved by noting that as x approaches infinity, xf(x) approaches L, which means that f(x) approaches L/x, which goes to zero as x goes to infinity since L is finite. This can also be stated in ε - M language as: for any ε > 0, there exists an integer M such that for all x > M, |f(x)| < ε.
  • #1
kingstrick
108
0

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.
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
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 .
 
  • #4
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??
 
  • #5
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.
 
  • #6
Correct :-)
 
  • #7
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?
 
  • #8
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, ...
 
  • #9
SammyS said:
Given an ε>0, there exists an integer, M, such for all x > M, ...

Thanks... I am an idiot!
 

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

1. What does it mean for a function to have a limit at infinity?

When a function has a limit at infinity, it means that as the input values of the function approach infinity, the output values approach a specific number. This number is known as the limit and is denoted by the letter "L".

2. How do you determine the limit at infinity of a function?

To determine the limit at infinity of a function, you can evaluate the function at increasingly large values of x. If the output values approach a specific number, then that number is the limit at infinity. Alternatively, you can use algebraic techniques such as factoring or rationalizing the denominator to simplify the function and identify the limit at infinity.

3. What does it mean if the limit at infinity of a function is equal to 0?

If the limit at infinity of a function is equal to 0, it means that as the input values of the function approach infinity, the output values approach 0. This can also be interpreted as the function approaching the x-axis as x goes to infinity.

4. Can a function have a limit at infinity at more than one point?

Yes, a function can have a limit at infinity at multiple points. This means that as x approaches infinity, the output values of the function approach a specific number at multiple points along the x-axis.

5. What is the relationship between a function's limit at infinity and its horizontal asymptote?

The limit at infinity of a function is equal to its horizontal asymptote. This means that as x approaches infinity, the output values of the function approach the value of the horizontal asymptote. Alternatively, if a function has a horizontal asymptote, then its limit at infinity is equal to the value of the asymptote.

Similar threads

Replies
13
Views
3K
Replies
2
Views
1K
Replies
6
Views
1K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
1
Views
992
Replies
2
Views
1K
Replies
4
Views
983
Back
Top