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I was curious about this exercise, because I thought it could be a valuable tool to use the theorem dell'asintoto ... I do not think that is the way, does anyone have any idea? !

Let $ h $ is a function defined on $(a, + \infty)$ and limited all intervals $(a, b)$

$a <$ b. Prove that if $$ \lim_ {x \to + \infty} [h (x +1)-h (x) ] = k, $$

then $$\lim_ {x \to + \infty} \frac {h (x)}{x} =k$$

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    "limited all intervals $(a,b)$"?2011-12-30
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    I interpret that to mean bounded on all intervals.2011-12-30

1 Answers 1

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Let $\epsilon>0$. There is an $N$ such that if $x>N$, then $|h(x+1)-h(x)-k|<\epsilon$. Therefore, $$|h(y+n)-h(y)-kn|=\left|\sum_{i=1}^{n}(h(y+i)-h(y+i-1))-k\right|\leq n\epsilon$$ for $y>N+1$.

Suppose that $A0$), we have

$$\frac{A-n\epsilon +(n-y)k}{y} < \frac{h(y)}{y}-k < \frac{B+n\epsilon +(n-y)k}{y}.$$

Since $y-n\in [N+1,N+2]$ and $\frac{n}{y}<\frac{y-N}{y}<1+\frac{|N|}{y}$, we therefore have

$$\left|\frac{h(y)}{y}-k\right| < \frac{\max(|A|,|B|)+|N|+2}{y}+\epsilon \left(1+\frac{|N|}{y}\right).$$

For large values of $y$, $\frac{\max(|A|,|B|)+|N|+2}y<\epsilon$ and $1+\frac{|N|}y<2$, so

$$\left|\frac{h(y)}{y}-k\right|<3\epsilon.$$

Since we can do this for every $\epsilon>0$, we must have $\displaystyle \lim_{x\to \infty} h(x)/x=k$.

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    In your second line, you mean $$|h(y+n) - h(y) - kn| = \left| \sum_{i=1}^n (h(y+i) - h(y+i-1) - k)\right| \le n \epsilon$$2011-12-30
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    @Robert Israel: Yes, thank you.2011-12-30