B Can f(x) and f'(x) both approach a non-zero constant?

[Saracen Rue]

Hello everyone! I've been curious about this for a while and couldn't come to a conclusion on my own so I've decided to ask it here.

I'm wondering if it's possible for a function, f(x), to have a rule which would allow it and it's derivative to both approach a constant value as x approaches infinity. For example, $lim_{(x→∞)} f(x) = 3$, $lim_{(x→∞)} f'(x) = 5$. Note, I don't mind if the constants are equal to each other, the only thing that's important is that they are non-zero constants.

Thank you all for your time :)

 

[Scolecites]

Hello everyone! I've been curious about this for a while and couldn't come to a conclusion on my own so I've decided to ask it here.

I'm wondering if it's possible for a function, f(x), to have a rule which would allow it and it's derivative to both approach a constant value as x approaches infinity. For example, $lim_{(x→∞)} f(x) = 3$, $lim_{(x→∞)} f'(x) = 5$. Note, I don't mind if the constants are equal to each other, the only thing that's important is that they are non-zero constants.

Thank you all for your time :)

Not that I can think of. The first derivative is a slope, implying that the function changes and would not approach a constant value. If f(x) were to approach a constant value, the slope of the function would be approaching zero for that domain.

 

[Saracen Rue]

Not that I can think of. The first derivative is a slope, implying that the function changes and would not approach a constant value. If f(x) were to approach a constant value, the slope of the function would be approaching zero for that domain.

Thank you for your advice. I had figured as much, but I still felt it necessary to ask in case there is a function that would act as an exception to the general rules that I don't yet know about.

 

[jostpuur]

You can use inequalities to prove the impossibility. If

<br /> \lim_{x\to\infty} f&#039;(x) = 5<br />

then there exists R such that f&#039;(x)&gt;4 for all x&gt;R. Then also

<br /> f(x) &gt; f(R)+ 4(x-R)<br />

holds for all x&gt;R

 

[SlowThinker]

How about something like 3+5*(x mod 1) ?
Say 3+5*(x mod (1/x)) ?
The limit is 3 and f'(x)=5, most of the time anyway.

 

[Saracen Rue]

How about something like 3+5*(x mod 1) ?
Say 3+5*(x mod (1/x)) ?
The limit is 3 and f'(x)=5, most of the time anyway.

That's very close to what I'm looking for! But not quite. As $x→∞$, $f(x)→8$. However $f'(x)$ only approaches $5$ as $x→-∞$. Is there a further step to manipulate this so that they both approach a constant as $x$ approaches positive infinity?

 

[SlowThinker]

As $x→∞$, $f(x)→8$. However $f'(x)$ only approaches $5$ as $x→-∞$.

Maybe your definition of "mod" is different from mine?
I'm pretty sure that f'(x) is exactly 5 everywhere, and also $3\leq f(x) \leq 3+5/x$ which approaches 3 as x goes to positive infinity.

 

[jostpuur]

It is not clear what you mean with that mod function. Can you write the same things in terms of the floor function x\mapsto \lfloor x\rfloor?

Anyway, it looks like you are speaking about functions that are not differentiable at all points. Why would you be interested in the limit

<br /> \lim_{x\to\infty} f&#039;(x)<br />

if f was not differentiable at least in some set ]R,\infty[ with some R\in\mathbb{R}?

 

[SlowThinker]

It is not clear what you mean with that mod function. Can you write the same things in terms of the floor function x\mapsto \lfloor x\rfloor?

$$f(x)=3+5(x^2-\lfloor x^2\rfloor)/x$$

Why would you be interested in the limit if f was not differentiable at least in some set ]R,\infty[ with some R\in\mathbb{R}?

It was the only way I could figure out to satisfy the requirement of the original poster.

 

[jostpuur]

It was the only way I could figure out to satisfy the requirement of the original poster.

It certainly is the only way to get a some kind of function requested by the original poster, since I already proved (or showed some relevant steps of a proof) that such function cannot exist if it is required to be differentiable for all sufficiently large x. The inequality in my proof can be justified by the mean value theorem, by the way, I forgot to mention it above.

 

[jostpuur]

I guess the original question has already been answered, but anyway, I could not help taking a look at the function you just defined, and I guess of course you have to keep doing some exercises to maintain your math skills

The formula

<br /> x\mapsto \frac{x^2 - \lfloor x^2\rfloor}{x}<br />

defines a function that has discontinuities at points x=\sqrt{1},\sqrt{2},\sqrt{3},\ldots, and is differentiable between these points. When x assumes values from an interval \sqrt{n}\leq x &lt; \sqrt{n+1}, the expression assumes values from the interval

<br /> 0\leq \frac{x^2 - \lfloor x^2\rfloor}{x} &lt; \frac{1}{\sqrt{n+1}}<br />

So these values are going to zero in the limit n\to\infty.

The derivative is

<br /> D_x \frac{x^2 - \lfloor x^2\rfloor}{x} = 1 + \frac{\lfloor x^2\rfloor}{x^2}<br />

When x assumes values from an interval \sqrt{n}&lt; x&lt;\sqrt{n+1}, the derivative assumes values from the interval

<br /> 1 + \frac{n}{n+1} &lt; 1 + \frac{\lfloor x^2\rfloor}{x^2} &lt; 2<br />

The lower bound can be written in the form

<br /> 2 - \frac{1}{n} + O\Big(\frac{1}{n^2}\Big)<br />

so we see that the values of the derivative get squeezed close to 2 at the limit n\to\infty.

 

[PeroK]

Another approach is to have a function that is not defined everywhere. E.g a function that is defined on a sequence of diminishing open intervals.

In this case you could have a function that is continuous and differentiable everywhere ( on its domain) but meets the criteria.

 

[SlowThinker]

The derivative is
<br /> D_x \frac{x^2 - \lfloor x^2\rfloor}{x} = 1 + \frac{\lfloor x^2\rfloor}{x^2}<br />

Are you sure about this? I might have made a mistake in rewriting x mod (1/x) into the floor function but I'm pretty sure that
$$\frac{d}{dx} (x\ \mod\ \text{anything}) = \frac{d}{dx} (x) = 1$$
except on points where the mod function does its thing.

Edit: Well you're right, it looks like something funny is going on here when there's a function of x in the denominator.

http://www.wolframalpha.com/input/?...mod+2,+x+mod+(1/2),+x+mod+(6/x)}+from+6+to+16