calculus - Does, $mathop {lim }limits_{x to +infty } f'(x) = + infty Leftrightarrow mathop {lim }limits_{x to +infty } frac{{f(x)}}{x} = + infty$?

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.

Answer

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.