Function and its derivatives at infinity Let $ f: \mathbb{R} \to \mathbb{R}$ be a doubly differentiable function. Suppose that $ lim_{x \to \infty} f(x) = 0 $ and $ lim_{x \to \infty} f''(x) = 0 $. Does this imply that $ lim_{x \to \infty} f'(x) = 0 $?
 A: Barbalat's lemma (proof here): 

If $f$ converges to some $L$ and $f'$ is uniformly continuous over some interval $[a,\infty)$, then $f'$ converges to $0$.

Since $f''$ converges (the value of the limit is irrelevant), $f''$ is bounded over some $[A,\infty)$, thus $f'$ is uniformly continuous over $[A,\infty)$.
A: Choose $\epsilon >0$, then there exists $N \in \mathbb{N}$ so that for all $y \geq N$ we have $|f(y)|, 2|f''(y)|< \epsilon/3$. By the Mean Value theorem for any $n \geq N$ there exists $n < a_n < n+1$ so that $$f'(a_n) = f(n+1) - f(n).$$ Let $x \in \mathbb{R}$ be so that $x >N+1$ and choose $n \in \mathbb{N}$ so that $0 < x-a_n \leq 2$, then by the Mean Value theorem there exists $a_n < b(n,x) < x$ so that $$f''(b(n,x)) = \frac{f'(x) - f'(a_n)}{x-a_n} = \frac{f'(x) - (f(n+1)-f(n))}{x-a_n}.$$ We can rearrange this to give $$f'(x) = f''(b(n,x))(x-a_n) + f(n+1) - f(n).$$ We note that as $$N \leq n <  a_n < b_n$$ this gives $$|f'(x)| \leq |f''(b(n,x))||(x-a_n)| + |f(n+1)| + |f(n)| < \epsilon/3 + \epsilon/3 +\epsilon/3 = \epsilon$$ for the desired result.
