1. Jul 10, 2010

### zetafunction

can exist an smooth function with the property

$$y(\infty) =0$$ and $$y'(\infty) =1$$ ?

the inverse case, a function that tends to 1 for big x and whose derivative tends to 0 is quite obvious but this case i am not sure if there will exist

2. Jul 10, 2010

### CompuChip

I suppose that you mean
$$\lim_{x \to \infty} y(x) = 0$$ and $$\lim_{x \to \infty} y'(x) = 1$$ ?

Actually, I think that for a smooth function to have a limit at infinity, the derivative should have limit 0 (at least that's what my intuition tells me: for the function to have a limit at infinity, it should become progressively more flat, so it doesn't run away from its limit value).

I have some other work now, but I will try to prove that rigorously later (if you want, give it a try yourself).

3. Jul 10, 2010

### losiu99

It follows easily from MVT.

4. Jul 10, 2010

### CompuChip

That's what I figured, but I got caught up in epsilons and deltas on the back of my scrap piece of paper.
After finishing my Saturday's to-do list I will take a completely blank paper of normal size and try again :)

5. Jul 10, 2010

### losiu99

On the second thought, derivative doesn't have to go to zero, unfortunetely. Consider $$\frac{\sin (x^2)}{x}$$. It clearly tends to zero, yet the derivative oscilates. Still, the derivative cannot tend to a nonzero number, and this follows from MVT for sure
Sorry for the mistake.

Edit: derivative of $$\frac{\sin (x^2)}{\sqrt{x}}$$ oscilates unboundedly, while the function goes to zero.

Last edited: Jul 10, 2010