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.

Let's suppose $f$ a continuous function where $$\lim_{x\to+\infty} { f(x+1)- f(x)}=l.$$ How to prove that $$\lim_{x\to +\infty}\frac{f(x)}{x}=l\,?$$

I've already proove by Cesaro theorem that $$\lim_{n\to +\infty}\frac{f(n) }{n}=l,$$ where $n$ is an integer. How to continue it?

share|improve this question
1  
I think this was asked before, recently... –  David Mitra Jan 3 '12 at 21:07
    
Doesn't it immediately follow from the limit for integers, and the fact that every $x$ lies between two integers? –  Scaramouche Jan 3 '12 at 21:25
    
@Scaramouche Define $f(x)$ to equal $0$ when $x$ is an integer and equal $1$, otherwise. The limit over the integers is $0$, but the limit over the reals does not exist. –  Austin Mohr Jan 3 '12 at 21:32
1  
Does the limit always exist? I think there's counter example for which the limit doesn't exist. –  bonnnnn2010 Jan 3 '12 at 22:01
2  

1 Answer 1

up vote 2 down vote accepted

Put $g(x)=f(x)-lx$. Then $g$ is continuous, $\lim_{x\to +\infty}g(x+1)-g(x)=0$ and we have to show that $\lim_{x\to+\infty}\frac{g(x)}x=0$. Fix $\varepsilon>0$, and $n_0$ an integer such that $|g(x+1)-g(x)|\leq\varepsilon$ if $x\geq n_0$. Let $M:=\sup_{x\in [x_0,x_0+1]}|g(x)|$, which is finite since $g$ is continuous. Let $x\geq n_0$. We can find $N=N(x)$ such that $x-N\in [x_0,x_0+1[$. Since $$g(x)-g(x-N)=\sum_{k=0}^{n-1}g(x-k)-g(x-k-1),$$ we have $\frac{|g(x)-g(x-N)|}N\leq\varepsilon$, hence $|g(x)-g(x-N)|\leq N\varepsilon$. We get $$\left|\frac{g(x)}x\right|\leq \frac{N\varepsilon}x+\frac{|g(x-N)|}x\leq \frac{N\varepsilon}x+\frac Mx,$$ and since $x_0\leq x-N$, $\frac{x_0}x\leq 1-\frac Nx$, so $\frac Nx\leq 1-\frac{x_0}x\leq 1$ and we got $$\forall \varepsilon>0,\,\forall x\geq n_0,\quad \left|\frac{g(x)}x\right|\leq \varepsilon+\frac Mx,$$ so for all $\varepsilon>0,\: \limsup_{x\to+\infty}\left|\frac{g(x)}x\right|\leq\varepsilon$, which is the wanted result.

share|improve this answer

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.