Is it true that f'(x) --> 0 as x --> oo?

Question

Suppose f is differentiable on [0, oo) and lim (x --> oo) f(x) exists. Does this implies that lim (x --> oo) f'(x) = 0? Apparently the answer is yes, because f has an horizontal assyntote. But I couldn't prove it formally.

Answer 1

No, it isn't always true. Consider the counterexample f(x) = [sin (x + 1)^2] / (x + 1), which is defined everywhere on the interval [0, ∞) f(x) is confined to the envelope y = ±1 / (x + 1) which squeezes arbitrarily close to zero as x goes to infinity. Therefore, f(x) has a limiting value.

However, when you differentiate this function, you get f'(x) = -sin ((x+1)^2) / (x+1)^2 + 2 cos ((x+1)^2). Although the second term remains bounded between - 2 and 2 as x approaches infinity, the amplitude of this term remains constant and does not go to zero. If the argument of the sine function is raised to a power higher than 2, the derivative isn't even bounded as x approaches infinity but the limit of the function value still goes to zero.

Edit: I added the constant to my argument for the counterexample to move the singularity at x = 0 out of the domain of interest. With f (x) = sin (x^2) / x, you have the side issue of showing that the derivative at f(0) exists.

Answer 2

No, this is not true. Consider some oscillation function which amplitude decreases with x, while period decreases at the same or higher rate. For example, let's take sin(x^2)/x. Its derivative is 2x*cos(x^2)/x - sin(x^2)/(x^2)=2cos(x^2)-sin(x^2)/(x^2). As you can see f(x)->0 while f'(x) does not. If you take function like sin(x^3)/x you will get even diverging derivative.

Answer 3

f(x) --> L as x --> infinity
f'(x) =0 since L is a constant at x = infinity.