Prove that $\lim_{x\to\infty}f''(x)=0$ if $\lim_{x\to\infty}f(x)=T$ and $\lim_{x\to\infty}f'''(x)=0$. Suppose that $f$ is a real function such that $f(x) \to T$, where $T$ is a finite limit, and that $f''' \to 0$ as $x \to \infty$.
Prove that $f''(x) \to 0$ as $x \to \infty$.
 A: Using Taylor-Lagrange formula:
$$f(x+1)=f(x)+f'(x)+\frac{1}{2}f''(x)+\frac{1}{3!}f'''(c_x)$$ where $c_x\in (x,x+1)$, and 
$$f(x-1)=f(x)-f'(x)+\frac{1}{2}f''(x)-\frac{1}{3!}f'''(c_x')$$ where $c_x'\in (x-1,x)$, we obtain 
$$f(x+1)+f(x-1)=2f(x)+f''(x)+\frac{1}{3!}f'''(c_x)-\frac{1}{3!}f'''(c_x')$$
so 
$$f''(x)=f(x+1)+f(x-1)-2f(x)-\frac{1}{3!}f'''(c_x)+\frac{1}{3!}f'''(c_x')$$
since $x\leq c_x\leq x+1$ and $x-1\leq c_x'\leq x$, then $c_x\to +\infty$ and $c_x'\to +\infty $ (as $x\to +\infty$). It follow that 
$$f''(x)=f(x+1)+f(x-1)-2f(x)-\frac{1}{3!}f'''(c_x)+\frac{1}{3!}f'''(c_x')\to T+T-2T+0-0=0$$
A: You can also show the result (or a generalization) using the Mean Value Theorem as your only tool:
Lemma 1. Let $g\colon (a,\infty)$ be a differentiable function such that each interval $(x_1,x_2)$ of length $> L$ contains a point $\xi$ with $|g(\xi)|<c$. Then each interval of length $> 2L+2$ contains a point $\xi$ where $|g'(\xi)|<c$.
Proof.
Let $(x_1,x_2)\subset (a,\infty)$ be an interval of length $> 2L$.
By assumption, there are $\xi_1\in(x_1,\frac{x_1+x_2}2-1)$ and $\xi_2\in(\frac{x_1+x_2}2+1,x_2)$ with $|g(\xi_{1,2})|<c$. By the MVT, ther exists $\xi\in(\xi_1,\xi_2)$ with $g'(\xi)=\frac{g(\xi_2)-g(\xi_1)}{\xi_2-\xi_1}$. Because $\xi_2-\xi_1>2$, we conclude $|g'(\xi)|<c$. $_\square$
Lemma 2. Let $f\colon\mathbb R\to\mathbb R$ be a function that is $n(\ge1)$ times differentiable and such that $T:=\lim_{x\to+\infty}f(x)$  and $U:=\lim_{x\to+\infty}f^{(n)}(x)$ exist. Then $U=0$ and, if $n>1$, $\lim_{x\to\infty}f^{(n-1)}(x)=0$.
Proof.
Let $\epsilon>0$ be given. Then there exists $a$ such that $|f(x)-T|<\frac\epsilon2$ and $|f^{(n)}(x)-U|<\frac{\epsilon}{2^{n+1}}$ for all $x\in(a,\infty)$. We apply lemma 1 to $f-T$ and its derivatives to show by induction that for $1\le k\le n$ each interval of length $>2^{k+1}-2$ in $(a,\infty)$ contains a point $\xi$ where $|f^{(k)}(\xi)|<\frac\epsilon2$. This immediately gives us $U=0$ (of course).
Now assume $n>1$ and let $x\in(a,\infty)$. As just shown, there exists $\xi\in(x,x+2^{n})$ with $|f^{(n-1)}(\xi)|<\frac\epsilon2$. Using the MVT again, there exists $\eta\in(x,\xi)$ with $f^{(n)}(\eta)=\frac{f^{(n-1)}(\xi)-f^{(n-1)}(x))}{\xi-x}$. Then 
$$|f^{(n-1)}(x)|\le |\xi-x|\cdot |f^{(n)}(\eta)|+|f^{(n-1)}(\xi)|< 2^n\cdot \frac\epsilon{2^{n+1}}+\frac\epsilon 2=\epsilon.$$
We have shown that $|f^{(n-1)}(x)|<\epsilon$ for all $x>a$, i.e. $\lim_{x\to+\infty}f^{(n-1)}(x)=0$. $_\square$
Corallary.  Let $f\colon\mathbb R\to\mathbb R$ be a function that is $n(\ge1)$ times differentiable and such that $\lim_{x\to+\infty}f(x)$  and $\lim_{x\to+\infty}f^{(n)}(x)$ exist. Then $\lim_{x\to +\infty}f^{(k)}(x)=0$ for $1\le k\le n$.
Proof. Use lemma 2 for downward induction. $_\square$
