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If $\lim\limits_{x→∞} f(x)$ and $\lim\limits_{x→∞} f''(x)$ both exist, then $\lim\limits_{x→∞} f''(x) = 0.$

You may use the fact that $\lim\limits_{x→∞} f(x)$and $\lim\limits_{x→∞} f'(x)$ both exist, then $\lim\limits_{x→∞} f'(x) = 0.$ Proof

I feel like this should be a simple application, but I'm not seeing a direct link.

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Without loss of generality we can have $\lim f'' \ge 0$; otherwise consider $-f$ instead.

Assume, for a contradiction, that $\lim f''(x)\to 2a>0$. Then there is $N$ such that $f''(x)>a$ for all $x\ge N$. But then $f'(x)\ge 1$ whenever $x>N+\frac{|f'(N)|+1}{a}$. How can $f$ then have a finite limit? It can't.

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