3
$\begingroup$

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?

  • 2
    http://math.stackexchange.com/questions/95079/asymptote-problem2012-01-03

1 Answers 1

2

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.