Suppose $f: \mathbb{R} \to \mathbb{R}$ is twice differentiable. Show that $\lim_{x \to \infty} f''(x) = 0$, given that $\lim_{x \to \infty} f(x)$ and $\lim_{x \to \infty} f''(x)$ exist.
My attempt:
Let $x > 0$. By MVT, there is $c_1(x) \in (x,2x)$ and $c_2(x) \in (3x,4x)$ with
$$1/x(f(2x)-f(x)) = f'(c_1(x)); \quad 1/x(f(4x)-f(3x)) = f'(c_2(x))$$
Again, by MVT, there is $c_3(x) \in (c_1(x), c_2(x))$ with
$$f''(c_3(x)) ( c_2(x)-c_1(x)) = f'(c_2(x))-f'(c_1(x)) = 1/x (f(2x)-f(x)-f(4x)+f(3x))$$
Taking $\lim_{x \to \infty}$ of both sides, we find:
$$\lim_{x \to \infty} f'' (c_3(x)) (c_2(x)-c_1(x)) = 0$$
Because $c_2(x) - c_1(x) > 3x - 2x = x \to \infty$, it must be the case that $\lim_{x \to \infty} f''(c_3(x)) = 0$
But $c_3(x) > c_1(x) > x \to \infty$, which implies that $\lim_{x \to \infty} f''(x) = 0$.
Is this correct?
Your approach is correct. Here is another approach:
Using this question we see that $f'(x) \to 0$ as $x\to\infty $. And using mean value theorem we have $f'(x+1)-f'(x)=f''(\xi)$ which shows that $f''(x) \to 0$.