If $\lim_{x \to \infty}f(x)$ exist then $\lim_{x \to \infty}f′(x)=0$ 
Let $f: [0, \infty) \longrightarrow \mathbb{R}$, where $\lim_{x \to \infty}f(x)$ exist, show that $\lim_{x \to \infty}f′(x)=0$

This fact is clearly intuitive to me but I could not write a rigorous demonstration of this.
 A: The fact is false.  Here is a counterexample: 
You can smooth the corners so that $f'(x)$ is defined everywhere, but alternates between $+1$ and $-1$, so has no limit as $x\to \infty$.  Meanwhile, $f(x)$ is squeezed between curves that approach zero, so $f(x)\to 0$.
A: This seems intuitive at first glance, but it's not true.  Consider the following function:
$$f(x) = \frac{\sin(x^2)}{x+1}$$

To make this easier to see, consider instead the corresponding example for $g:(0, \infty) \rightarrow \mathbb{R}$:
$$g(x) = \frac{\sin(x^2)}{x}$$
And computing a derivative:
$$g'(x) = 2\cos(x^2) - \frac{\sin(x^2)}{x^2}$$
Certainly, $\displaystyle \lim_{x \rightarrow \infty} g(x) = 0$, but $\displaystyle \lim_{x \rightarrow \infty} g'(x)$ does not exist.  
The above example was built out of this one so that it will be defined at $x = 0$.
A: First I think the necessary assumption is $f'(x)$ exist for all $x\in[0,\infty)$ otherwise the conterexamples are given in other answers. Here is another one, $f(x)=\frac{1}{x}$ when $x$ is rational and $0$ when $x$ is irrational, the function is discontinuous everywhere(therefore no derivative exist) but satisfies your criteria.
Assume $f'(x)$ exist everywhere, let $lim_{x\rightarrow\infty}f'(x)=a>0$. We can prove $lim_{x\rightarrow\infty}f(x)=\infty$, for given $\epsilon=\frac{a}{2}>0$, we can find $\delta>0$ such that $x>\delta$ implies $f'(x)>a-\epsilon=\frac{a}{2}>0$. We know continuous function($f'$ exist so f continuous) is integrable and we have $$f(\delta+\gamma)=f(\delta)+\int_{\delta}^{\delta+\gamma}f'(x)dx>f(\delta)+\frac{a\gamma}{2}$$ for any $\gamma>0$. This implies $f(x)$ can be arbitrary large for large enough $x$, and that's what $lim_{x\rightarrow\infty}f(x)=\infty$ means.
The prove of the cases $a<0$ or $a=\pm\infty$ are similar.
