Let $f:[0,\infty]\to R$ be differentiable on $(0, \infty)$, and $f'(x)\to b$ as $x \to \infty$. Show that $\lim_{x \to \infty}\frac{f(x)}{x}=b$

Question

This is actually part (c) of the original question. Part (a) asks to prove for any $h>0$, we have $\lim_{x\to\infty}\frac{f(x+h)-f(x)}{h}=b$. Part (b) asks to prove if $f(x) \to a$ as $x\to\infty$, then $b=0$. I know how to do (a)(b) using Mean Value Theorem. I think I should use Mean Value Theorem here. For any $x>0$, there exists $c\in (0, x)$ such that $f'(c)=\frac{f(x)-f(0)}{x}$. But as $x\to\infty$, $c$ may not approches to $\infty$. I don't know how part (c) is related to (a)(b). I can't figure out whether I should use the trick of add and subtract or others.

Answer 1

I'll start by doing the two first parts, then I'll edit the post for the last.

For the first part, you must use Rolle's theorem : for every $x$ and $h>0$, there is $\theta_x \in ]x,x+h[$ such that $\frac{1}{h}(f(x+h) -f(x)) = f'(\theta_x)$. Choosing$\epsilon$ small and $x$ great enough, we have $|f'(\theta_x) - b|< \epsilon$, so it is clear that whenever $x \to \infty$, $$\frac{1}{h}(f(x+h) -f(x)) \to b$$

The second part should now be clear, for if we fix $h>0$ and take $x$ great enough to ensure that $|f(y)|<\frac{h\epsilon}{2}$ whenever $y>x$, then we have for every $y>x$ the following inequality : $$|\frac{1}{h}(f(y+h) - f(y))|\leq \frac{1}{h}(h\epsilon) = \epsilon$$

So the quantity $\frac{1}{h}(f(x+h) -f(x))$ goes to $0$ as $x \to \infty$, and by the uniqueness of the limit, $b=0$.

Edit for the answer. Someone dit it from scratch, but I'll try to do it using part 2 of your exercise. Suppose that $f$ has a limit $a \in \mathbb{R}$. Then, by part 2 we must have $b = 0$ and it also clear that, $f$ being bounded, $\frac{f(x)}{x}$ goes to $0$ as $x \to \infty$. So in this case we have $\frac{f(x)}{x} \to b$.

Answer 2

Pick $\epsilon>0$. Then there exists $x_0$ such that $|f'(x)-b|<\frac\epsilon3$ for $x>x_0$. Then by MWT for $x>x_0$ there exists $\xi\in(x_0,x)$ with $f(x)-f(x_0)=(x-x_0)f'(\xi)$. Hence $$\frac{f(x)}{x}-b=\frac{f(x_0)}{x}+f'(\xi)-b-\frac{x_0f'(\xi)}x $$ We already have $ |f'(\xi)-b|<\frac\epsilon3$. For $x>\frac{3|f(x_0)|}\epsilon$ we also have $\left|\frac{f(x_0)}{x}\right|<\frac\epsilon 3$. Finally, for $x>\frac{3|x_0|(|b|+\frac\epsilon3)}{\epsilon}$, also $\left|\frac{x_0f'(\xi)}{x}\right|<\frac\epsilon3$. That is, for $x$ large enough we have $$\left|\frac{f(x)}x-b\right|<\epsilon$$