Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a twice differentiable function such that $\lim_{x\to\infty}f(x)=0$ and $f''$ is bounded.

Question

Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a twice differentiable function such that $\lim_{x\to\infty}f(x)=0$ and $f''$ is bounded. Prove that $\lim_{x\to\infty}f'(x)=0$

I am trying to use L'Hospital rule. Please give some hints.

You just need $f'$ to be uniformly continuous for this to hold. — copper.hat, Jun 19 '22 at 18:45

score 0 · Answer 1 · answered Jun 19 '22 at 18:16

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I don't know how you could use L'Hopitals rule for this since it requires: $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)=\pm\infty$$ for $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$ otherwise you could say: $$\lim_{x\to\infty}f'(x)=\lim_{x\to\infty}\frac{f'(x)}1=\lim_{x\to\infty}\frac{f(x)}{x}=0$$

Obviously for a function to tend to a particular value its derivative must tend to 0.

answered Jun 19 '22 at 18:16

Henry Lee

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I do not think that your last sentence is correct. The existence of $\lim_{x\to\infty}f(x)$ does not imply that $\lim_{x\to\infty}f'(x) = 0$. – Martin R Jun 19 '22 at 18:23
I agree with this. – urt43as Jun 19 '22 at 18:23

Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a twice differentiable function such that $\lim_{x\to\infty}f(x)=0$ and $f''$ is bounded.

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