- #1

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Is there a theorem that says it's derivative, f'(x), also approaches 0 as x goes to infinity?

Thanks.

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- Thread starter vincent_vega
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- #1

- 32

- 0

Is there a theorem that says it's derivative, f'(x), also approaches 0 as x goes to infinity?

Thanks.

- #2

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[tex]f(x)=\frac{\sin(x^2)}{x}[/tex]

What is true, that if [itex]f^\prime(x)[/itex] has a limit, then this limit must be 0.

- #3

Bacle2

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f(b)-f(a)=f'(c)(b-a) . Maybe you can partition [0,oo)into [0,1],[0,2],...,[n,n+1].

Then:

f(1)=f(0)+f'(co) ;f(2)=f(1)+f'(c1) ;.....f(n)=f(n-1)+f'(c_(n-1)).

Then you can find a closed form for f(n).

- #4

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micromass's example shows this isn't going to work.If f is differentiable in (-oo,oo) , use the mean value theorem:

- #5

Bacle2

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Of course I'm assuming f'(x) is defined as x-->oo , that is implied in my argument.

Basically, take [0,b] . Then

f(b)-f(0)=f'(c)(b)

If f' is defined everywhere and we let b-->oo , then the limit cannot be 0 unless f'(c) decreases to zero.

If the OP says "its derivative approaches 0 as x --> infinity" seems to me to assume that the derivative is defined as x-->oo.

Basically, take [0,b] . Then

f(b)-f(0)=f'(c)(b)

If f' is defined everywhere and we let b-->oo , then the limit cannot be 0 unless f'(c) decreases to zero.

If the OP says "its derivative approaches 0 as x --> infinity" seems to me to assume that the derivative is defined as x-->oo.

Last edited:

- #6

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You mean, that the limit is defined? If your argument relies on that then it needs to be stated.Of course I'm assuming f'(x) is defined as x-->oo , that is implied in my argument.

- #7

Bacle2

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This looks to me like an assumption that f'(x) is defined as x-->oo

- #8

Bacle2

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Maybe the OP can clarify the conditions of the problem to eliminate ambiguity?

- #9

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No, that's what he's trying to prove. He didn't say "if f' approaches a limit, that limit is 0", so the most reasonable interpretation is that he wants to prove "f' approaches a limit, and that limit is 0".From the OP:" its derivative, f'(x), also approaches 0 as x goes to infinity? "

This looks to me like an assumption that f'(x) is defined as x-->oo

- #10

Bacle2

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no reason to assume it exists in a specific subset of the real line only, nor reason

to assume otherwise. In your interpretation, why didn't the OP say something like

is f'(x) defined, and if so , what is its limit. He refers to f'(x) which states that f'(x)

exists. It may exist somewhere or everywhere.

The problem is posed sloppily ; I think out of basic manners, the OP should clarify.

- #11

pwsnafu

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f'(x) is said to exist without any qualification

Then it does not follow that the limit ##\lim_{x\rightarrow\infty}f'(x)## is defined. The statement "f' exists" is limited to ##x \in \mathbb{R}## because the domain of f is the real numbers and not the extended reals. What happens as ##x\rightarrow\infty## is considered a separate condition, and must explicitly be mentioned.

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