Proving limit of f(x), f'(x) and f"(x) as x approaches infinity

[songoku]

Homework Statement

Please see below

Relevant Equations

Limit
Derivative from 1st Principle
Asymptotes
Taylor Series
L'Hopital rule

I imagine $f(x)$ has horizontal asymptote at $x=k$. Since the graph of $f(x)$ will be close to horizontal as $x \rightarrow \infty$, the slope of the graph will be close to zero so $\lim_{x \rightarrow \infty} f'(x) = \lim_{x \rightarrow \infty} f^{"} (x) = 0$

But how to put it in mathematical way? Thanks

 

[PeroK]

Homework Statement:: Please see below
Relevant Equations:: Limit
Derivative from 1st Principle
Asymptotes
Taylor Series
L'Hopital rule

View attachment 314940

I imagine $f(x)$ has horizontal asymptote at $x=k$. Since the graph of $f(x)$ will be close to horizontal as $x \rightarrow \infty$, the slope of the graph will be close to zero so $\lim_{x \rightarrow \infty} f'(x) = \lim_{x \rightarrow \infty} f^{"} (x) = 0$

But how to put it in mathematical way? Thanks

The first thing I'd do is look for a counterexample. Did you think of that?

I have no idea why this should be true and I don't believe it. That's my starting point.

 

[Orodruin]

Homework Statement:: Please see below
Relevant Equations:: Limit
Derivative from 1st Principle
Asymptotes
Taylor Series
L'Hopital rule

View attachment 314940

I imagine $f(x)$ has horizontal asymptote at $x=k$. Since the graph of $f(x)$ will be close to horizontal as $x \rightarrow \infty$, the slope of the graph will be close to zero so $\lim_{x \rightarrow \infty} f'(x) = \lim_{x \rightarrow \infty} f^{"} (x) = 0$

But how to put it in mathematical way? Thanks

Do you mean $y = k$? You cannot have $x = k$ and let $x \to \infty$ at the same time.

Edit: And yes, as @PeroK hints at - you should first make sure that there are no obvious counter examples before you set out to prove something.

 

[PeroK]

Edit: And yes, as @PeroK hints at - you should first make sure that there are no obvious counter examples before you set out to prove something.

I was thinking that looking for a counterexample would lead to an insight into why the statement is true.

 

[Orodruin]

I was thinking that looking for a counterexample would lead to an insight into why the statement is true.

It is not though, since counter examples exist.

Edit: To be clear, this statement is about what OP wrote, not the question posed in the attached picture.

 

[songoku]

Do you mean $y = k$? You cannot have $x = k$ and let $x \to \infty$ at the same time.

Yes, sorry I mean $y=k$

The first thing I'd do is look for a counterexample. Did you think of that?

I have no idea why this should be true and I don't believe it. That's my starting point.

Edit: And yes, as @PeroK hints at - you should first make sure that there are no obvious counter examples before you set out to prove something.

What I did was thinking which function is possible to have horizontal asymptote, I think about rational function and exponential function and those 2 functions fulfill the conditions in the question.

As for counterexample, nothing come to my mind to prove the question is wrong and also to prove my statement in OP is wrong. I think about polynomial graph, trigonometry, square root, logarithm but none of them satisfy $\lim_{x \rightarrow +\infty}=k$

Thanks

 

[PeroK]

What about $f(x) = \frac 1 x \sin(x^2)$?

 

[renormalize]

As for counterexample, nothing come to my mind to prove the question is wrong and also to prove my statement in OP is wrong. I think about polynomial graph, trigonometry, square root, logarithm but none of them satisfy $\lim_{x \rightarrow +\infty}=k$

How about $\lim_{x\rightarrow\infty}\arctan\left(x\right)=\frac{\pi}{2}$? The arctan function has the "Taylor series at infinity" (technically an asymptotic expansion): $$\arctan\left(x\right)=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^{3}}-\frac{1}{5x^{5}}+\ldots$$ Can you apply this same idea to your function $f(x)$?

 

[Orodruin]

I would also note that the question you are asked in the picture is very different from your statement in the OP.

 

[songoku]

What about $f(x) = \frac 1 x \sin(x^2)$?

For this:
$$\lim_{x \rightarrow \infty} f(x) = 0$$
$$f'(x)=\frac{2x^2 \cos(x^2)-\sin(x^2)}{x^2}$$
and $\lim_{x \rightarrow \infty} f'(x)$ does not exist

Both do not satisfy the conditions given by the question. Is this supposed to be the counterexample?

How about $\lim_{x\rightarrow\infty}\arctan\left(x\right)=\frac{\pi}{2}$? The arctan function has the "Taylor series at infinity" (technically an asymptotic expansion): $$\arctan\left(x\right)=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^{3}}-\frac{1}{5x^{5}}+\ldots$$ Can you apply this same idea to your function $f(x)$?

Sorry I don't understand the hint. $f(x)=\arctan (x)$ satisfies all the conditions given in the question but how to use that to prove the question?

Do you mean I need to find Taylor expansion for $f(x)$? But how to do that if I don't know $f(x)$?

I would also note that the question you are asked in the picture is very different from your statement in the OP.

You mean the question asks to prove $\lim_{x \rightarrow \infty} f(x)=k$ while I assume it is true without proving it?

Thanks

 

[Orodruin]

Both do not satisfy the conditions given by the question. Is this supposed to be the counterexample?

It is a counter example to your statement in the OP where you are trying to prove that the limit of f’(x) is zero if the limit of f(x) is k.

You mean the question asks to prove limx→∞f(x)=k while I assume it is true without proving it?

Yes, you need to prove it, not assume it. Furthermore, the condition given by the question is more restrictive than your assumption.

 

[PeroK]

For this:
$$\lim_{x \rightarrow \infty} f(x) = 0$$
$$f'(x)=\frac{2x^2 \cos(x^2)-\sin(x^2)}{x^2}$$
and $\lim_{x \rightarrow \infty} f'(x)$ does not exist

Both do not satisfy the conditions given by the question. Is this supposed to be the counterexample?

First, I have to say that this problem may be significantly beyond your abilities at this stage. I think you don't even understand what the question is asking. We could try a simpler question to try to explain what you are being asked.

Suppose we have a function $g(x)$ such that $\lim_{x \to \infty} g(x) = a$, where $a$ is a real number. I.e. the limit exists and is finite. Now, suppose that $\lim_{x \to \infty} g'(x) = b$, where $b$ is also a real number. Then we can prove that $b = 0$. That might be something you could try to prove.

However, it's not necessarily the case that $\lim_{x \to \infty} g'(x)$ exists. The above function is an example of this.

Now, here's the idea. Could we find a function $g(x)$ such that neither $\lim_{x \to \infty} g(x)$ nor $\lim_{x \to \infty} g'(x)$ exist and yet: $\lim_{x \to \infty} (g(x) + g'(x))$ exists and is finite?

That might be an easier problem to think about. In other words, either prove that:

If $$\lim_{x \to \infty} (g(x) + g'(x)) = k$$ then $$\lim_{x \to \infty} g(x) = k \ \ \text{and} \ \ \lim_{x \to \infty} g'(x) = 0$$Or, find a counterexample.

 

[songoku]

It is a counter example to your statement in the OP where you are trying to prove that the limit of f’(x) is zero if the limit of f(x) is k.Yes, you need to prove it, not assume it. Furthermore, the condition given by the question is more restrictive than your assumption.

Ah ok

First, I have to say that this problem may be significantly beyond your abilities at this stage. I think you don't even understand what the question is asking. We could try a simpler question to try to explain what you are being asked.

I also have the same thought. I don't even know what is the starting point to solve this question. L'Hopital can not be used because the form is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$, Taylor expansion can not be used so I tried derivative from the first principle and got:
$$\lim_{x \rightarrow \infty} (f(x)+2f'(x)+f"(x)=k$$
$$\lim_{x \rightarrow \infty} \left(f(x)+2 \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}+\lim_{h \rightarrow 0} \frac{f'(x+h)-f'(x)}{h} \right)=k$$

But I don't think this is the correct approach.

Let me rephrase the question:
Prove that if $\lim_{x \rightarrow \infty} (f(x)+2f'(x)+f"(x)=k$ and $\lim_{x \rightarrow \infty} f(x) \neq 0$ then $\lim_{x \rightarrow \infty} f(x)=k$ , $\lim_{x \rightarrow \infty} f'(x)=0$ and $\lim_{x \rightarrow \infty} f"(x)=0$

Do I interpret the question correctly?

We could try a simpler question to try to explain what you are being asked.

Suppose we have a function $g(x)$ such that $\lim_{x \to \infty} g(x) = a$, where $a$ is a real number. I.e. the limit exists and is finite. Now, suppose that $\lim_{x \to \infty} g'(x) = b$, where $b$ is also a real number. Then we can prove that $b = 0$. That might be something you could try to prove.

For this one, the question is:
Prove that if $\lim_{x \to \infty} g(x) = a$ and $\lim_{x \to \infty} g'(x) = b$, where $b$ is a real number, then $b = 0$

My attempt:
$$\frac{d}{dx} \left(\lim_{x \to \infty} g(x) \right) = \frac{d}{dx} (a)$$
$$\lim_{x \rightarrow \infty} \frac{d}{dx} g(x) = 0$$
$$\lim_{x \rightarrow \infty} g'(x) = 0$$

Now, here's the idea. Could we find a function $g(x)$ such that neither $\lim_{x \to \infty} g(x)$ nor $\lim_{x \to \infty} g'(x)$ exist and yet: $\lim_{x \to \infty} (g(x) + g'(x))$ exists and is finite?

I am not sure, I can't think of any. I will try again

Thanks

 

[PeroK]

Ah ok

I also have the same thought. I don't even know what is the starting point to solve this question. L'Hopital can not be used because the form is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$, Taylor expansion can not be used so I tried derivative from the first principle and got:
$$\lim_{x \rightarrow \infty} (f(x)+2f'(x)+f"(x)=k$$
$$\lim_{x \rightarrow \infty} \left(f(x)+2 \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}+\lim_{h \rightarrow 0} \frac{f'(x+h)-f'(x)}{h} \right)=k$$

But I don't think this is the correct approach.

I agree. That doesn't look promising.

Let me rephrase the question:
Prove that if $\lim_{x \rightarrow \infty} (f(x)+2f'(x)+f"(x)=k$ and $\lim_{x \rightarrow \infty} f(x) \neq 0$ then $\lim_{x \rightarrow \infty} f(x)=k$ , $\lim_{x \rightarrow \infty} f'(x)=0$ and $\lim_{x \rightarrow \infty} f"(x)=0$

Do I interpret the question correctly?

Yes.

For this one, the question is:
Prove that if $\lim_{x \to \infty} g(x) = a$ and $\lim_{x \to \infty} g'(x) = b$, where $b$ is a real number, then $b = 0$

My attempt:
$$\frac{d}{dx} \left(\lim_{x \to \infty} g(x) \right) = \frac{d}{dx} (a)$$
$$\lim_{x \rightarrow \infty} \frac{d}{dx} g(x) = 0$$
$$\lim_{x \rightarrow \infty} g'(x) = 0$$

You need something more formal than that. The basic idea is that if $g'(x) \to b$, where $b \ne 0$, then $g(x)$ must eventually be changing steadily and cannot itself tend to a finite limit. You can formalise this using the $\epsilon-\delta$ definition.

I am not sure, I can't think of any. I will try again

The question looks very tricky to me. I haven't spotted a solution yet.

 

[Orodruin]

$$\frac{d}{dx} \left(\lim_{x \to \infty} g(x) \right) = \frac{d}{dx} (a)$$

This statement is nonsensical. You cannot differentiate with respect to x after taking the limit $x \to \infty$.

 

[pasmith]

You can obtain examples of functions satisfying $f''(x)+ 2f'(x) + f(x) \to k$ by solving the (linear, constant coefficient) ODE $$f'' + 2f' + f = g(x)$$ where $g$ is any function such that $g(x) \to k$. Thus you can obtain examples of monotonic tendency to $k$ such as $$g(x) = k(1- e^{-x}), \qquad f(x) = Ae^{-x} + Bxe^{-x} + k(1 - \tfrac12x^2e^{-x})$$ or oscillatory tendency such as $$g(x) = k + e^{-x}\sin x,\qquad f(x) = Ae^{-x} + Bxe^{-x} + k - e^{-x}\sin x$$ where setting $A = B = 0$ gives an $f$ which does not have a horizontal asymptote.

In answering the actual question, it may assist you to make use of the fundamental theorem and a basic property of integrals that if $g(x) < h(x)$ for all $x > R$ then for all $x > R$, $$\int_R^x g(t)\,dt < \int_R^x h(t)\,dt$$ in order to find a lower bound for $f$ which tends to $+\infty$ if it is given that $f'(x) \to b > 0$, and to find an upper bound for $f$ which tends to $-\infty$ if it is given that $f'(x) \to b < 0$.

 

[songoku]

The question looks very tricky to me. I haven't spotted a solution yet.

You mean the question I posted in OP?

In answering the actual question, it may assist you to make use of the fundamental theorem and a basic property of integrals that if $g(x) < h(x)$ for all $x > R$ then for all $x > R$, $$\int_R^x g(t)\,dt < \int_R^x h(t)\,dt$$ in order to find a lower bound for $f$ which tends to $+\infty$ if it is given that $f'(x) \to b > 0$, and to find an upper bound for $f$ which tends to $-\infty$ if it is given that $f'(x) \to b < 0$.

Sorry I think I need to solve the question without using integral. Is there an alternative approach?

Thanks

 

[renormalize]

songoku, it turns out that the desired proof can be written in just a few lines using an old trick due to mathematician G.H. Hardy, namely: multiply and divide $f(x)$ by $e^{x}$ and then use L'Hospital's rule to evaluate the limit.

Here are some hints to get you started: observe that $f(x)+2f^{\prime}(x)+f^{\prime\prime}(x)=\left(e^{x}f(x)\right)^{\prime\prime}/e^{x}$ and note that the limit can be rewritten as $\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\frac{e^{x}f(x)}{e^{x}}$. Keeping the first hint in mind, apply L'Hospital's rule twice to the right side and take the limit. Repeat the same trick with only one use of L'Hospital's rule and calculate the limit again. What can you conclude from the two expressions you get for this limit?

Note that what you will have proved is a version of Barbalat's Lemma:

(https://en.wikipedia.org/wiki/Lyapunov_stability)

 

[songoku]

songoku, it turns out that the desired proof can be written in just a few lines using an old trick due to mathematician G.H. Hardy, namely: multiply and divide $f(x)$ by $e^{x}$ and then use L'Hospital's rule to evaluate the limit.

Here are some hints to get you started: observe that $f(x)+2f^{\prime}(x)+f^{\prime\prime}(x)=\left(e^{x}f(x)\right)^{\prime\prime}/e^{x}$ and note that the limit can be rewritten as $\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\frac{e^{x}f(x)}{e^{x}}$. Keeping the first hint in mind, apply L'Hospital's rule twice to the right side and take the limit. Repeat the same trick with only one use of L'Hospital's rule and calculate the limit again. What can you conclude from the two expressions you get for this limit?

Note that what you will have proved is a version of Barbalat's Lemma:

View attachment 315052
(https://en.wikipedia.org/wiki/Lyapunov_stability)

I think I understand the hint. But there is no way I can think this by myself during the test

Thank you very much for the help and explanation PeroK, Orodruin, renormalize, pasmith

 

[Orodruin]

I mean, the pattern of note here is the coefficients 1-2-1, which should be recognisable as the binomial coefficients of of 2c2, 2c1, 2c0. Because of this, you can kind of sort of get at rewriting the expression as something that includes $f(x)$ and a function that is directly proportional to its own derivative. In particular, it holds that
$$
\frac{d^n}{dx^n}(f(x)g(x)) = \sum_{k = 0}^n {n \choose k} f^{(k)}(x) g^{(n-k)}(x)
$$
and so, if $g'(x) = \alpha g(x)$ (i.e., $g(x) = e^{\alpha x}$) then
$$
\frac{d^n}{dx^n}(f(x)g(x)) = g(x) \sum_{k=0}^n {n \choose k} \alpha^{n-k}f^{(k)}(x).
$$
In particular, for $n = 2$ you would have
$$
\frac{(f(x)g(x))''}{g(x)} = \alpha^2 f(x) + 2\alpha f'(x) + f''(x),
$$
which is equal to the original expression if $\alpha = 1$ (just keeping $\alpha$ general to note that other possibilities exist).

However, the big question is: Can you complete the proof with the hint that was given?

 

[songoku]

I mean, the pattern of note here is the coefficients 1-2-1, which should be recognisable as the binomial coefficients of of 2c2, 2c1, 2c0. Because of this, you can kind of sort of get at rewriting the expression as something that includes $f(x)$ and a function that is directly proportional to its own derivative. In particular, it holds that
$$
\frac{d^n}{dx^n}(f(x)g(x)) = \sum_{k = 0}^n {n \choose k} f^{(k)}(x) g^{(n-k)}(x)
$$
and so, if $g'(x) = \alpha g(x)$ (i.e., $g(x) = e^{\alpha x}$) then
$$
\frac{d^n}{dx^n}(f(x)g(x)) = g(x) \sum_{k=0}^n {n \choose k} \alpha^{n-k}f^{(k)}(x).
$$
In particular, for $n = 2$ you would have
$$
\frac{(f(x)g(x))''}{g(x)} = \alpha^2 f(x) + 2\alpha f'(x) + f''(x),
$$
which is equal to the original expression if $\alpha = 1$ (just keeping $\alpha$ general to note that other possibilities exist).

However, the big question is: Can you complete the proof with the hint that was given?

I can try.

$$\lim_{x \rightarrow \infty} (f(x)+2f'(x)+f^{"} (x)=k$$
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})^{"}}{e^{x}}=k$$
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})^{"}}{(e^{x})^{"}}=k$$

Since it is an indeterminate form, I am thinking to backtrack so:
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})^{"}}{(e^{x})^{"}}=k$$
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})^{'}}{(e^{x})^{'}}=k$$
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})}{(e^{x})}=k$$

Is this valid?

Thanks

 

[WWGD]

I can try.

$$\lim_{x \rightarrow \infty} (f(x)+2f'(x)+f^{"} (x)=k$$
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})^{"}}{e^{x}}=k$$
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})^{"}}{(e^{x})^{"}}=k$$

Since it is an indeterminate form, I am thinking to backtrack so:
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})^{"}}{(e^{x})^{"}}=k$$
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})^{'}}{(e^{x})^{'}}=k$$
$$\lim_{x \rightarrow \infty} \frac{(f(x).e^{x})}{(e^{x})}=k$$

Is this valid?

Thanks

Of the bat, it seems your expression on the left would be a function of $x$. But let me think it further. And it would imply:
$\frac{(f(x)e^x)'}{f(x)g(x)}=1$, so that $(f(x)e^x)'=f(x)g(x)$.Then the expression $f(x)e^x$ must equal $e^x$, as $e^x$ is the only solution to $h'(x)=h(x)$, ultimately
Edit: I did not consider the limits here.

 

[WWGD]

Maybe this could move things forward:
Try the mvt with a Telescope:
Find the least integer a' larger than a. Then use
$\Sigma f(a+1)-f(a)=\Sigma f'(a_*) ; a_*\in (a, a+1)$ , etc.( Since $f$ is assumed differentliable in $( a, \infty))$That would show , by properties of Telescopes, since f' is finite, thst the nth term goes to 0. A similar idea for

 

[renormalize]

I can try.

You're attacking this in the wrong direction; just proceed methodically from left to right:
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\frac{e^{x}f(x)}{e^{x}}=\lim_{x\rightarrow\infty}\frac{\left(e^{x}f(x)\right)^{\prime\prime}}{\left(e^{x}\right)^{\prime\prime}}=\lim_{x\rightarrow\infty}\frac{\left(e^{x}f(x)\right)^{\prime\prime}}{e^{x}}=\lim_{x\rightarrow\infty}\left(f(x)+2f^{\prime}(x)+f^{\prime\prime}(x)\right)=k$$where I've applied L'Hospitals rule twice in the middle step. This establishes that $\lim_{x\rightarrow\infty}f(x)=k$ and $\lim_{x\rightarrow\infty}\left(2f^{\prime}(x)+f^{\prime\prime}(x)\right)=0$.

Can you now repeat this line of reasoning, but with only one use of L'Hospital's rule, to ultimately conclude that the limits of $f^{\prime}$ and $f^{\prime\prime}$ must each vanish individually, thereby completing the proof?

 

[Orodruin]

Of the bat, it seems your expression on the left would be a function of x.

This is not the case. The limit $x \to \infty$ is taken.

 

[songoku]

You're attacking this in the wrong direction; just proceed methodically from left to right:
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\frac{e^{x}f(x)}{e^{x}}=\lim_{x\rightarrow\infty}\frac{\left(e^{x}f(x)\right)^{\prime\prime}}{\left(e^{x}\right)^{\prime\prime}}=\lim_{x\rightarrow\infty}\frac{\left(e^{x}f(x)\right)^{\prime\prime}}{e^{x}}=\lim_{x\rightarrow\infty}\left(f(x)+2f^{\prime}(x)+f^{\prime\prime}(x)\right)=k$$where I've applied L'Hospitals rule twice in the middle step. This establishes that $\lim_{x\rightarrow\infty}f(x)=k$ and $\lim_{x\rightarrow\infty}\left(2f^{\prime}(x)+f^{\prime\prime}(x)\right)=0$.

Can you now repeat this line of reasoning, but with only one use of L'Hospital's rule, to ultimately conclude that the limits of $f^{\prime}$ and $f^{\prime\prime}$ must each vanish individually, thereby completing the proof?

I can.

Thank you very much for the help and explanation Orodruin, renormalize, WWGD

 

[PeroK]

You're attacking this in the wrong direction; just proceed methodically from left to right:
$$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\frac{e^{x}f(x)}{e^{x}}=\lim_{x\rightarrow\infty}\frac{\left(e^{x}f(x)\right)^{\prime\prime}}{\left(e^{x}\right)^{\prime\prime}}=\lim_{x\rightarrow\infty}\frac{\left(e^{x}f(x)\right)^{\prime\prime}}{e^{x}}=\lim_{x\rightarrow\infty}\left(f(x)+2f^{\prime}(x)+f^{\prime\prime}(x)\right)=k$$where I've applied L'Hospitals rule twice in the middle step. This establishes that $\lim_{x\rightarrow\infty}f(x)=k$ and $\lim_{x\rightarrow\infty}\left(2f^{\prime}(x)+f^{\prime\prime}(x)\right)=0$.

Can you now repeat this line of reasoning, but with only one use of L'Hospital's rule, to ultimately conclude that the limits of $f^{\prime}$ and $f^{\prime\prime}$ must each vanish individually, thereby completing the proof?

There's something important missing there. Note that one of the criteria in the original problem statement has not been used.

 

[renormalize]

There's something important missing there. Note that one of the criteria in the original problem statement has not been used.

Are you referring here to the condition $\lim_{x\rightarrow\infty}f(x)\neq0$?

If so, I think this requirement is included in the original problem simply to justify the use of L'Hospital's rule on $\lim_{x\rightarrow\infty}\frac{e^{x}f(x)}{e^{x}}$. Since $e^{x}\rightarrow\infty$ as $x\rightarrow\infty$, then so must $e^{x}f(x)$ in order that the ratio $\frac{e^{x}f(x)}{e^{x}}$ approach the indeterminant form $\frac{\infty}{\infty}$ that can be evaluated by L'Hospital's rule. On the contrary, if we allow $\lim_{x\rightarrow\infty}f(x)=0$, we could choose $f$ to be a function like $e^{-2x}$ giving $\frac{e^{x}f(x)}{e^{x}}=\frac{e^{-x}}{e^{x}}\rightarrow\frac{0}{\infty}$, precluding the use of L'Hospital's rule and spoiling the method of proof.

Even so, by introducing the function $F(x)=f(x)-L$ where $L=\lim_{x\rightarrow\infty}f(x)$, it's clear from the original proof that the "edge case" $\lim_{x\rightarrow\infty}F(x)=0$ still implies that $\lim_{x\rightarrow\infty}F^{\prime}(x)=\lim_{x\rightarrow\infty}F^{\prime\prime}(x)=\ldots=0$.

 

[PeroK]

Are you referring here to the condition $\lim_{x\rightarrow\infty}f(x)\neq0$?

If so, I think this requirement is included in the original problem simply to justify the use of L'Hospital's rule on $\lim_{x\rightarrow\infty}\frac{e^{x}f(x)}{e^{x}}$. Since $e^{x}\rightarrow\infty$ as $x\rightarrow\infty$, then so must $e^{x}f(x)$ in order that the ratio $\frac{e^{x}f(x)}{e^{x}}$ approach the indeterminant form $\frac{\infty}{\infty}$ that can be evaluated by L'Hospital's rule.

Yes, but what if $f(x)$ oscillates and $\lim_{x \to \infty} f(x)$ does not exist? I think the proof here requires that $\lim_{x \to \infty} f(x)$ exists.

 

[Office_Shredder]

Yes, but what if $f(x)$ oscillates and $\lim_{x \to \infty} f(x)$ does not exist? I think the proof here requires that $\lim_{x \to \infty} f(x)$ exists.

This bothered me for a while. For this l'hopital proof you need the limit to exist. But is that a flaw of the proof, or does a true counterexample exist if we don't have this condition?

It turns out the condition is not needed. This is probably the finest piece of analysis I've done in my life. if I've missed a more trivial proof, please do not tell me lol.

Lemma: if $\lim_{x\to \infty}$ does not exist, and $f$ is continuous, then either the limit is $\pm \infty$, or there exists $k$ and $\epsilon$ such that for all $M>0$, there exists $x_+>M$ and $x_- > M$ such that $f(x_+)>k+\epsilon$ and $f(x_-)<k-\epsilon$. In other words, if the function doesn't diverge to infinity, then it must bounce back and forth across at least one value repeatedly.

Proof:

Spoiler

We will say a set is unbounded if it's unbounded on the right, i.e. it contains values that are arbitrarily large and positive. let $A$ be the set of all $y$ such that $\{x : f(x)>y\}$ is unbounded. This is an interval of the form $(-\infty, a|$ for some $a$ (possibly infinity) - if $y\in A$, then $z\in A$ for all $z<y$. I make no claim whether the right endpoint of this interval is contained in it, hence the $|$ instead of $)$ or $]$. Similarly $B$ is the set of all $y$ such that $\{x:f(x)<y\}$ is unbounded - $B=|b,\infty)$ for some $b$. If $a<b$, then for $a<k<b$, $k\notin B$ which means there is some $M_1$ such that for $x>M_1$#, $f(x)\geq k$. Similarly $x\notin A$ so for some $M_2$, $x>M_2$ means $f(x) \leq k$. Hence for $x>\max(M_2,M_2)$ we have $f(x)=k$. But there are many such $k$ so this is impossible.

If $a>b$ we are done. Pick any $k$ and $\epsilon >0$ such that $a<k-\epsilon$ and $k+\epsilon < b$.

if $a=b$, there are three possibilities - they are both $\pm \infty$, or they are both the same number. Let's suppose $a=b=\infty$. Then $B$ is empty - for all $y$, there exists $M$ such that $x>M$ means $f(x)>y$. So $\lim_{x\to \infty}f(x)=\infty$. Similarly if $a=b=-\infty$ the limit is $-\infty$.

if $a=b$ is a number we will call $k$, we know for all $y>k$, there is some $M$ such that $f(x)<y$ for $x>M$. Similarly for any $y<k$, $f(x)>y$ for all $x$ large enough. Hence $\lim_{x\to \infty}f(x)=k$.

this covers all the cases. $\lim_{x\to \infty} f(x)$ either exists, is $\pm \infty$, or there is some $k$ and $\epsilon$ such that $f(x)>k+\epsilon$ and $f(x)<k-\epsilon$ for unbounded sets of $x$

Lemma: if $f(x)$ is differentiable, and $\lim_{x\to \infty}f(x)+ f'(x)=k$ exists, then $lim_{x\to \infty} f(x)=k$ and $\lim_{x\to \infty} f'(x)=0$

Proof:

Spoiler

We apply the first lemma to $f(x)$. If $\lim_{x\to \infty} f(x)=\infty$, then for any $M>0$ if $x$ is large enough, $f(x)>M$. We also know that there is some $x'>x$ such that $f(x')>f(x)$ since the limit is infinity. But then there is some point $x<x''<x'$ such that $f'(x'')>0$ by the mean value theorem. Then $f(x'')+f'(x'')>M$. We can do this for arbitrary $M$ and for arbitrarily large $x''$, so the limit $\lim_{x\to \infty}f(x)+f'(x)$ cannot exist. It might be $\infty$, but as a side note it doesn't have to - $f(x)=x^2$ and then modify the graph by occasionally dropping really fast for a very short interval is a counterexample.

So if $\lim_{x\to \infty} f(x)+f'(x)$ exists, we know $\lim_{x\to \infty} f(x)$ cannot be infinity. Similarly it can't be negative infinity.

This means either the limit exists, or there is some $p$ for which $f$ oscillates back and forth getting at least $\epsilon$ away each time

If we have the oscillating behavior - by the intermediate value theorem on the points that are at least $\epsilon$ away we have an infinite sequence $x_n$ of values for which $f(x_n)=p$, and on each $[x_n,x_{n+1}]$, we know $|f(x)-p|>\epsilon$ for at least one $x$. Let's assume that we are on an interval where we know for some $x$ $f(x)>p+\epsilon$ (similar argument for the other intervals) We know since $f$ is continuous it achieves a maximum on $[x_n,x_{n+1}]$ and the maximum is not on the boundary, since the maximum has $f(x)>p+\epsilon$ and $f(x_n)=f(x_{n+1})=p$.

Since the extreme point is in the interior of the interval, it must have a derivative equal to zero, so we have found a point for which $f(x)+f'(x)>p+\epsilon$.

hence we have an unbounded sequence of points for which $f(x)+f'(x)$ is either larger than $p×\epsilon$ or smaller than $p-\epsilon$, and therefore $\lim_{x\to \infty} f(x)+f'(x)$ doesn't exist in this case.

so the only possible case left is $\lim_{x\to \infty} f(x)= p$. Then $\lim_{x\to \infty} f'(x)=k-p$. If $p\neq k$, let's say $k>p$, then there is some $M$ such that $x>M$ means $f'(x)>(k-p)/2$. Then for $x>M$, $f(x)-f(0)=\int_0^M f'(t)dt + \int_M^x f'(t)dt > C+ (x-M)(k-p)/2$ for some constant $C=f(M)-f(0)$. So $f(x)$ is bounded from below by a linear function of positive slope, contradicting the claim that $\lim_{x\to \infty}f(x)$ is not infinity.

Of course, if $k=p$ then we get the claim of the lemma as desired

Then the final proof is: Let $g(x)= f(x)+f'(x)$. If $\lim_{x\to \infty} f(x)+2f'(x)+f''(x)=k$ for any real $k$, then $f(x)+2f'(x)+f''(x)=g(x)+g'(x)$. Hence by the proceeding lemma, $\lim_{x\to \infty} g(x)=k$ and $\lim_{x\to \infty} g'(x)=0$. So $\lim_{x\to \infty} f(x)+f'(x)=k$. Again by the proceeding lemma, $\lim_{x\to \infty}f(x)=k$ and $lim_{x\to \infty}f'(x)=0$. Subtracting these out of the original limit gives $\lim_{x\to \infty}f''x(x)=0$ as well, which finishes the proof.

 

[WWGD]

Maybe this could move things forward:
Try the mvt with a Telescope:
Find the least integer a' larger than a. Then use
$\Sigma f(a+1)-f(a)=\Sigma f'(a_*) ; a_*\in (a, a+1)$ , etc.( Since $f$ is assumed differentliable in $( a, \infty))$That would show , by properties of Telescopes, since f' is finite, thst the nth term goes to 0. A similar idea for

I'd appreciate a basis for your doubt Is it not the case that

$Lim _{x \rightarrow \infty} \Sigma_{j=1}^{\infty} (a_{j+1} -a_j) = a_1 - lim_{ j \rightarrow \infty} a_j$,
which in our case equals $f(1)- Lim_{ n \rightarrow \infty} f'(x_n)?$, which converges iff the sequence $f_n:= f'(x_n) ; x_n:=(a +n)$ converges, or do you oppose it on a different basis?

 

[Office_Shredder]

@WWGD that only constructs a single unbounded sequence of points for which the derivative converges to zero. How do you handle the rest of them?

 

[WWGD]

@WWGD that only constructs a single unbounded sequence of points for which the derivative converges to zero. How do you handle the rest of them?

I agree; it was just an idea for a path ahead to show $Lim_{x \infty}f'(x)=0$. I'm working on the details, using sequential continuity , given $f \in C^2(a, \infty)$. But maybe I should have been explicit that it was not intented as a worked solution.