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It's known that:

$$\lim \limits_{x \to \infty} \space f({x})=0$$

$$|f''(x)| < M$$

How do I prove that:

$$\lim \limits_{x \to \infty} \space f'({x})=0$$

I tried using the mean value theorem, but I don't really know how to approach the question.

Paul
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1 Answers1

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I didn't see this one in the duplicates..

Proof by contradiction. Suppose not, and there is $\epsilon > 0$ and $x_n \rightarrow \infty$ such that $|f'(x_n)| > \epsilon$. Since $|f''(x)| < M$, for $x$ in the interval $[x_n, x_n + {\epsilon \over 2M}]$ one has $|f'(x)| > {\epsilon \over 2}$. So by the mean value theorem, $|f(x_n) - f(x_n + {\epsilon \over 2M})| > {\epsilon \over 2}\times {\epsilon \over 2M} = {\epsilon^2 \over 4M}$. Hence at least one of $|f(x_n)|$ and $|f(x_n + {\epsilon \over 2M})|$ is greater than ${\epsilon^2 \over 8M}$. Letting $n$ go to infinity contradicts the assumption that $\lim_{x \rightarrow \infty} f(x) = 0$.

Zarrax
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  • Best one I've seen! (But this is robjohn's argument from my second link above.) – David Mitra Jul 08 '14 at 17:11
  • Same basic idea yes.. but I think this is more efficient so I'll keep it here for people to see. – Zarrax Jul 08 '14 at 17:15
  • Thanks. Could you clarify this: Since |f′′(x)|<M, for x in the interval [xn,xn+ϵ2M] one has |f′(x)|>ϵ2. – Paul Jul 08 '14 at 17:48
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    By Taylor (or mean value theorem), $f'(x) = f'(x_n) + f''(y)(x - x_n)$ for some $y$ in the interval. So $|f'(x)| > |f'(x_n)| - |f''(y)(x - x_n)| > \epsilon - M|x-x_n| > \epsilon - M{\epsilon \over 2M} = {\epsilon \over 2}$ – Zarrax Jul 08 '14 at 17:56