Question:
True or False.
Let $f:\mathbb{R} \mapsto \mathbb{R}$ a differentiable function s.t. $\lim_{x \rightarrow \infty } f'(x) = l' > 0$ ($l'$ positive and finite). So: $\lim_{x \rightarrow \infty } f(x) = \infty$
Answer:
I answer that this is true and below my proof:
1 – By definition $\lim_{x \rightarrow \infty } f'(x) = l' > 0$ implies that if we choose $\epsilon = l' / 2 >0$, $\exists x_0$ s.t. $\forall x>x_0$ it will be verify that $f'(x) \in (l'-\epsilon; l'+\epsilon)=(l'/2;3l'/2) \Rightarrow 0<l'/2<f'(x)<3l'/2$. So by the first derivative test $f(x)$ is stricly increasing on the open interval $(x_0, \infty)$.
2 – By absurd let suppose that $\lim_{x \rightarrow \infty } f(x) = l < \infty$.
Let remember that $f(x)$ is differentiable $\forall x$. It means that for all $x$ the following expression is define $f'(x) = \lim_{h \rightarrow 0 } \frac{f(x+h)-f(x)}{h}$ . But as $x$ goes to infinity we will get: $\lim_{x \rightarrow \infty}f(x+h)=l=\lim_{x \rightarrow \infty}f(x)$ for every $h>0$ we can choose. Hence $f'(x) = 0$ and by assumption it is not possible.
(In a shorter way we could writte: $\lim_{x \rightarrow \infty } f'(x) = \lim_{x \rightarrow \infty } \lim_{h \rightarrow 0 } \frac{f(x+h)-f(x)}{h}=\frac{l-l}{h}=0$)
Q.E.D
Is this correct?
Best Answer
You can argue by mean value theorem: Since $\lim_{x \rightarrow \infty} f'(x) = l'>0$, for any $\epsilon>0$, we have for $y>B_{\epsilon}>0$, $f'(y) > l'-\epsilon$. Let $x$ be such that $x > B_{\epsilon}$ and $f(x)< \infty$. If could not pick such $x$, $f(x) = \infty$, $\forall, x>B_{\epsilon}$ and we are done. By mean value theorem, for every $y>x$, there is a $\psi \in [x,y]$, we have $f(y) - f(x) = f'(\psi) (y-x) > (l'-\epsilon) (y-x)$, for all $y>x>B_{\epsilon}$. Now fix $x$ and increase $y$, we have that $\lim_{y \rightarrow \infty} f(y)-f(x) > (l'-\epsilon) \lim_{y \rightarrow \infty} (y-x) = \infty$. Hence $\lim_{y \rightarrow \infty} f(y) = \infty$.