how to prove this question about derivative and differentiation Let 
$$
f:\mathbb{R}\to \mathbb{R}
$$ 
such that $f ',f'',f'''$ exist and $\lim_{x\to+\infty} f(x)=t$ exists if  $  \lim_{x\to+\infty} f'''(x)=0$. 
Then prove that
$$
\lim_{x\to+\infty} f'(x) = \lim_{x\to+\infty} f''(x)=0.
$$
Thanks in advance 
 A: This solution is inspired by Aaron Sun's attempt, and using the same hypotheses: let $r>0$ be fixed. By Taylor's formula we have, for all $x$:$$f(x+r)=f(x)+f^\prime(x)r+f^{\prime\prime}(x)\,\frac{r^2}2+f^{\prime\prime\prime}(\xi)\frac{r^3}6\,,$$
where $\xi=\xi(x,r)$ lies between $x$ and $x+r$. When $x\to\infty$, the same happens to $x+r$ and $\xi$. Taking this limit into the previous equality and simplifying we get $$\lim_{x\to\infty}2f^\prime(x)+rf^{\prime\prime}(x)=0$$ (recall that $r$ is fixed). Now we take two different values of $r$, say $r=1,2$, from which we obtain $$\lim_{x\to\infty}f^{\prime\prime}(x)=\bigl(\lim_{x\to\infty}2f^\prime(x)+2f^{\prime\prime}(x)\bigl)-\bigl(\lim_{x\to\infty}2f^\prime(x)+f^{\prime\prime}(x)\bigl)=0-0=0\,,$$
from which we easily conclude that $\lim_{x\to\infty}f^\prime(x)=0$ as well.
A: I assume that you mean "$\lim_{x\to+\infty} f(x)=t$ exists and  $  \lim_{x\to+\infty} f'''(x)=0$".
Using Taylor Theorem ,we can express $f(x)$ like this:$$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(\xi)}{6}(x-a)^3,$$ which $a$ is a real constant, and $\xi$ is between $x$ and $a$.
Let $x=a+1$,we obtain $$f(a+1)=f(a)+f'(a)+\frac{f''(a)}{2}+\frac{f'''(\xi)}{6},$$which $a<\xi<a+1$.Then,let $a\to+\infty$,we obtain $t=t+\lim_{x\to+\infty}[f'(x)+f''(x)/2]$,i.e.,$$\lim_{x\to+\infty}[2f'(x)+f''(x)]=0.$$
Then,similarly,let $x=a-1$ and $a\to+\infty$,we obtain $$\lim_{x\to+\infty}[-2f'(x)+f''(x)]=0.$$
So,$$\lim_{x\to+\infty}[2f'(x)+f''(x)]+\lim_{x\to+\infty}[-2f'(x)+f''(x)]=2\lim_{x\to+\infty}f''(x)=0,$$and$$\lim_{x\to+\infty}[2f'(x)+f''(x)]-\lim_{x\to+\infty}f''(x)=2\lim_{x\to+\infty}f'(x)=0.$$
