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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?

  • How can you prove that $\frac{f(2x)-f(x)-f(4x)+f(3x)}{x}\to 0?$ – mfl Sep 05 '18 at 20:06
  • $f(2x) \to L$ for some L by assumption. Hence, $f(2x)/x \to 0$. Similarly for the other terms. I'm more concerned about the last step though. –  Sep 05 '18 at 20:06
  • OK. One of the conditions is that $\lim_{x \to \infty} f(x)$ exists. Is it a real number or is it allowed to be $\infty?$ – mfl Sep 05 '18 at 20:08
  • Exists means real number for me. –  Sep 05 '18 at 20:09
  • I think the line where I deduce $\lim_{x \to \infty} f''(x)$ exists from the existence of $\lim_{x \to \infty} f"(c_3(x))$ is false. –  Sep 05 '18 at 20:11
  • It is not false. By hypothesis you have that $\lim_{x \to \infty} f''(x)$ exists. So if $\lim_{x \to \infty} f''(c_3(x))=0$ (since $c_3(x)\to \infty$) you get that $\lim_{x \to \infty} f''(x)=0.$ – mfl Sep 05 '18 at 20:14
  • Ah yes, I see! Does that mean that the proof is correct? –  Sep 05 '18 at 20:15
  • In my opinion it is correct. – mfl Sep 05 '18 at 20:15
  • @mfl Thank you for your time. –  Sep 05 '18 at 20:16
  • 1
    You're welcome. – mfl Sep 05 '18 at 20:16

1 Answers1

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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 $f''(x) \to 0$.

Paramanand Singh
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