Does, $\lim\limits_{x \to +\infty } f'(x) = + \infty \Leftrightarrow \lim\limits_{x \to +\infty } \frac{{f(x)}}{x} = + \infty $?
The "left to right" of the biconditional is true. As noted in another answer, we can use L'hopital. But I will utilize a direct approach. We need to show that for arbitrarily large $M$, we have for sufficiently large $x$ the inequality $\frac{f(x)}{x} > M$.
By assumption, for any arbitrarily large $M$ we have $f'(x) > 2M$ when $x>x_0$. This means $f(x) \geq f(x_0) + 2M(x-x_0)$ for $x > x_{0}$. Note also that there is an $x_1$ such that for all $x > x_1$, $f(x_0) + 2M(x-x_0) > {Mx}$. Hence, we can see that for $x > \max\{x_1, x_0\}$ we have $$\frac{f(x)}{x} \geq \frac{f(x_0) + 2M(x-x_0)}{x} > \frac{Mx}{x} = M $$
The "right to left" of the biconditional is false. Consider $f(x) = x^2(\sin x + 2)$. This is positive and bounded below by $x^2$, hence $\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty$ but $f'$ oscillates as $x \to +\infty$.
We can say something weaker, however, namely the following
Theorem: Let $f \in C^1(\mathbb{R})$ such that $$\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty$$ Then we have $$\limsup_{x \to +\infty} \ f'(x) = +\infty$$
To prove this, first note that for $f$, we can assume $f(0) = 0$ without any loss of generality. Indeed, define $g(x) = f(x) - f(0)$ and note $\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty \Longleftrightarrow \lim_{x \to +\infty} \frac{g(x)}{x} = +\infty$ and also $f' = g'$.
We can prove by contradiction. Suppose the $\lim \sup$ is finite or $-\infty$. This means $f'$ is bounded above in $[M, +\infty)$ for some $M>0$. Since $f'$ is continuous, by the extreme value theorem it is bounded above in $[0,M]$, and hence it is bounded above in $[0, +\infty)$. By the mean value theorem, we have that $\frac{f(x)}{x} = f'(\alpha)$ for some $\alpha$ in $[0, x]$. Letting $x \to +\infty$ we can see that $f'(\alpha)$ takes on arbitrarily large positive values, which contradicts the fact that $f'$ is bounded above in $[0, +\infty)$.
This can probably be modified so that the $C^1$ condition can be relaxed (e.g., to allow for cases where $f'$ is discontinuous), but I'm not sure how to do that.
- It is true that if $\lim_{x\to +\infty} f'(x)=+\infty$ then $\lim_{x\to +\infty} \frac{f(x)}{x} = + \infty$. This can be proved by using the methods given by Dr.MV's answer to your question.
- It is in general false that if $\lim_{x\to +\infty} \frac{f(x)}{x}=+\infty$ then $\lim_{x\to +\infty} f'(x)=+\infty$.
Counter-example: Let $f(x)=x^2 + \sin x^3$. Then $\lim_{x\to +\infty} \frac{f(x)}{x} =+\infty$. But $f'(x) = 2x+3x^2\cos x^3$. Note that $f'(x)$ is continuous for all $x\in \mathbf{R}$, but since the sign of $\cos x^3$ could vary as $x$ goes to $+\infty$, $\lim_{x\to +\infty}f'(x)$ doesn't exist and is not $+\infty$.
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.