Let $f(x):(0,\infty)\rightarrow \mathbb{R}$, $f$ is differentiable. It is given that: $\mathop {\lim }\limits_{x \to \infty } f'(x) = L > 0$.
Show $f$ is not bounded.
I've seen a proof which can be summarized:
$$\forall \varepsilon > 0\exists M.\forall x > M:\left| {f'(x) – L} \right| < \varepsilon $$
Particularly,
$L-1 < f'(x) < L+1$ where $\varepsilon=1$.
We argue by contradiction $f$ is bounded. Therefore, there are $M,n$ which are maximum and minimum values of $f$, respectively.
Defining: $A=M-m>0$.
At this point, we know that $\forall x>x_0: f'(x)>L-1$.
Let us choose $x_1>x_0$, such that: $$x_1-x_0 > {A \over L-1}$$
Long story short, if you use Lagrange's Mean Value Theorem you will find that:
$$\left| {f({x_1}) – f({x_0})} \right| > A = m – n$$
A contradiction, which implies $f$ is not bounded as requested.
Well, After reading the proof couple of times I think I get the intention.
To show $f$ is not bounded we'd like to use (by contradiction) maximum and minimum.
What is bothering me is the definition of $x_1-x_0 > {A \over L-1}$.
Sure, it did the work as you can see, but how can one think of choosing this inequality? I mean, what does it represent actually? I'm trying to interpret the meaning of the ratio on the righthand side relative to substraction on the lefthand side. And maybe earlier question, why choosing $\varepsilon=1$?
It is very important for me to understand it, in order to adapt the ability solving this kind of problems.
Thanks in advance.
Best Answer
Think of a line with slope $L-1$. That means that any change in $x$ will get multiplied by $L-1$ to get the change in $y$. Since we want the total change in $y$ to be more than $A$, that means that the change in $x$ must be more than $A/(L-1)$.
And of course, since $L-1$ is the minimum of $f'$, the total change could of course be larger; but this is a lower bound. (Note we need $L>1$ for our choice of $\epsilon$ to work, but you can always choose an appropriate $\epsilon$, say $L/2$ for whatever $L$ happens to be.)