Does the Limit of f'(x) Imply f''(x) Equals Zero?

[terhorst]

#34 on the much-discussed http://ftp.ets.org/pub/gre/Math.pdf" :

Suppose $$f$$ is a differentiable function with $$\lim\limits_{x \to \infty }f(x)=K$$ and $$\lim\limits_{x \to \infty }f'(x)=L$$ for some $$K,L$$ finite. Which must be true?

1. $$L=0$$
2. $$\lim\limits_{x \to \infty }f''(x)=0$$
3. $$K=L$$
4. $$f$$ is constant.
5. $$f'$$ is constant.

Answer is 1. Is this because $$f$$ might be $$C^1$$? Can you give an example of a function where the limit of the first derivative exists but the limit of the second derivative is not zero? Thanks!

 

[nicktacik]

I am a little befuddled by this. If 1. is true, it seems like 2. must also be true.
Let $$g(x)=f'(x)$$
We know
$$\lim_{x\rightarrow\infty} g(x) = K = 0$$
So it should follow that
$$\lim_{x\rightarrow\infty} g'(x) = \lim_{x\rightarrow\infty} f''(x) = L = 0$$

 

[Dick]

If the limit of the second derivative exists then it is zero. But it may not exist - even if the function is C^2. Try sin(x^2)/x^2.