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Problem: Prove that if $\lim_{x \to \infty} f(x)$ exists and $\lim_{x \to \infty} f''(x)$ exists, then $\lim_{x \to \infty} f''(x)=\lim_{x \to \infty} f'(x)=0$.

I just need help in proving that $\lim_{x \to \infty} f'(x)$ exists. Hints would also be appreciated

Subham Jaiswal
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  • you don't need to prove that, that is the basic assumption of this problem. – MoonKnight Jan 26 '16 at 17:39
  • $f(x)=x$ do not have limit at $x\rightarrow +\infty$ – EQJ Jan 26 '16 at 17:42
  • I am assuming that the limit is finite. – EQJ Jan 26 '16 at 17:42
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    @MoonKnight the problem just assume that the limits at infinite of $f(x)$ and $f''(x)$ exist but there is nothing about $f'(x)$. – EQJ Jan 26 '16 at 17:44
  • A more general version is that if $\lim_{x \to \infty}f(x)$ exists and $f''(x)$ is bounded for all $x > a$ then $f'(x) \to 0$ as $x \to \infty$. See http://math.stackexchange.com/q/730411/72031 The existence of limit of $f''(x)$ in this question ensures its boundedness for all $x$ greater than some number. – Paramanand Singh Jan 28 '16 at 04:10

1 Answers1

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By Taylor's theorem, for any $x $ there exists $x < \xi_x < x+1$ such that

$$f(x+1) = f(x) + f'(x) + f''(\xi_x)/2.$$

Then

$$\lim_{x \to \infty}f'(x) = \lim_{x \to \infty}f(x+1)- \lim_{x \to \infty}f(x)- \lim_{x \to \infty}f''(\xi_x)/2 = - \lim_{x \to \infty}f''(x)/2. $$

RRL
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