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

1. Mar 26, 2017

### 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 :)

2. Mar 26, 2017

### Scolecites

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.

3. Mar 26, 2017

### Saracen Rue

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.

4. Mar 26, 2017

### jostpuur

You can use inequalities to prove the impossibility. If

$$\lim_{x\to\infty} f'(x) = 5$$

then there exists $R$ such that $f'(x)>4$ for all $x>R$. Then also

$$f(x) > f(R)+ 4(x-R)$$

holds for all $x>R$

5. Mar 27, 2017

### 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.

6. Mar 27, 2017

### Saracen Rue

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?

7. Mar 27, 2017

### SlowThinker

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.

8. Mar 27, 2017

### 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

$$\lim_{x\to\infty} f'(x)$$

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

9. Mar 28, 2017

### SlowThinker

$$f(x)=3+5(x^2-\lfloor x^2\rfloor)/x$$
It was the only way I could figure out to satisfy the requirement of the original poster.

10. Mar 28, 2017

### jostpuur

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.

11. Mar 28, 2017

### 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

$$x\mapsto \frac{x^2 - \lfloor x^2\rfloor}{x}$$

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 $\sqrt{n}\leq x < \sqrt{n+1}$, the expression assumes values from the interval

$$0\leq \frac{x^2 - \lfloor x^2\rfloor}{x} < \frac{1}{\sqrt{n+1}}$$

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

The derivative is

$$D_x \frac{x^2 - \lfloor x^2\rfloor}{x} = 1 + \frac{\lfloor x^2\rfloor}{x^2}$$

When $x$ assumes values from an interval $\sqrt{n}< x<\sqrt{n+1}$, the derivative assumes values from the interval

$$1 + \frac{n}{n+1} < 1 + \frac{\lfloor x^2\rfloor}{x^2} < 2$$

The lower bound can be written in the form

$$2 - \frac{1}{n} + O\Big(\frac{1}{n^2}\Big)$$

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

12. Mar 28, 2017

### 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.

13. Mar 28, 2017

### SlowThinker

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

Last edited: Mar 28, 2017