Does, $\lim\limits_{x \to +\infty } f'(x) = + \infty \Leftrightarrow \lim\limits_{x \to +\infty } \frac{{f(x)}}{x} = + \infty $? Let $f:\Bbb R \to \Bbb R$ be a differentiable function. If $\mathop {\lim }\limits_{x \to  + \infty } \frac{{f(x)}}{x} =  + \infty $, it is always true that $\mathop {\lim }\limits_{x \to  + \infty } f'(x) =  + \infty $? How about the converse?
For example, $\mathop {\lim }\limits_{x \to  + \infty } \frac{{\ln x}}{x} = 0$ is finite, then we can see $\mathop {\lim }\limits_{x \to  + \infty } (\ln x)' = 0$ is finite. $\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2}}}{x} =  + \infty $ so $\mathop {\lim }\limits_{x \to  + \infty } ({x^2})' = \mathop {\lim }\limits_{x \to  + \infty } x =  + \infty $. So the claim seems good to me, but I don't know how to actually prove it. $\mathop {\lim }\limits_{x \to  + \infty } f'(x) = \mathop {\lim }\limits_{x \to \infty } \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$, I don't know how to deal with this mixed limit. Also since the limits in the proposition diverges, it looks like mean value theorem sort of thing cannot apply here.

If $\lim_{x\to \infty}f'(x)$ exists, then from L'Hospital's Rule we have
$$\lim_{x\to \infty}\frac{f(x)}{x}=\lim_{x\to \infty}f'(x)$$
regardless of whether $\lim_{x\to \infty}f(x)$ exists or not (See the note that follows Case 2 of THIS ARTICLE).
Hence, if $\lim_{x\to \infty}f'(x)=\infty$, then $\lim_{x\to \infty}\frac{f(x)}{x}=\infty$ also.

