Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a differentiable function on $(a,+\infty)$ such as $\lim \limits_{x \to\infty } \frac{f(x)}{x}=0$ prove the following: $$ \lim \limits_{x \to\infty } \inf |f'(x)|=0 $$

I just can't see how to do it... (even after understanding How to show that $\lim\limits_{x \to \infty} f'(x) = 0$ implies $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$?)

share|improve this question
1  
For some $\epsilon > 0,$ what happens if $f'(x) > \epsilon$ for all $x > X_0,$ where $X_0$ is some large positive number? The other case is, what happens if $f'(x) < - \epsilon$ with everything else written the same? –  Will Jagy Feb 25 '13 at 5:40
    
@WillJagy than f(x) is monotonically increasing to infinity, but it doesn't yet a contradict –  User Feb 25 '13 at 5:46
    
Anyway, there is still a little more to it after that, but it is a start. –  Will Jagy Feb 25 '13 at 5:46
    
User, yes, monotone to infinity, and what about $f(x) / x?$ –  Will Jagy Feb 25 '13 at 5:47
    
Or, what about $(f(x)-f(X_0)) / (x - X_0)?$ –  Will Jagy Feb 25 '13 at 5:54
show 3 more comments

2 Answers

up vote 4 down vote accepted

It suffices to show that there exists a sequence $x_n\to \infty$ such that $f'(x_n)\to 0$.

By the mean value theorem, for each $n$ there exists $x_n$ with $n<x_n<2n$ such that $$ f'(x_n) = \frac{f(2n)-f(n)}{2n-n}=\frac{f(2n)-f(n)}{n}=2\frac{f(2n)}{2n}-\frac{f(n)}{n}. $$ Since $\lim_{x\to \infty} f(x)/x=0$, it follows that the right hand side tends to zero, and hence $$ \lim_{n\to \infty} f'(x_n)=0. $$ By the squeeze theorem, the fact that $x_n>n$ implies that $$ \lim_{n\to \infty} x_n=\infty. $$ This finishes the proof.

share|improve this answer
add comment

In case $f'(x)>\epsilon$ for all $x\geq x_0$ then (show that) $f(x)>f(x_0)+\epsilon(x-x_0)$ for all $x\geq x_0$.
Then show that $f(x)>f(x_0)+\epsilon(x-x_0)>\frac{\epsilon}{2}x$ for all $x\geq x_1$ for some $x_1\geq x_0$...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.