For a non-negative integer $k$, $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l\implies \lim_{x\to \infty}\frac{f(x)}{x^{k+1}}=\frac l{k+1}$

Question

Suppose that $f$ defined on $(a,\infty)$ is bounded on each finite interval $(a,b),a>b$. For a non-negative integer $k$, it is to be shown that $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l\implies \lim_{x\to \infty}\frac{f(x)}{x^{k+1}}=\frac l{k+1}$

Given an $\displaystyle \epsilon >0,$there exists an integer $\displaystyle N$ such that

$$\displaystyle x \geq N \Longrightarrow \ \left| \frac{f( x+1) -f( x)}{x^{k}} -l \right| < \epsilon $$

Noting that \begin{align*} f( x) & =\sum _{i=1}^{[ x]} f( x-[ x] +i) -f( x-[ x] +i-1) +\color{red}{f( x-[ x])} \end{align*} it follows that \begin{align*} f( x) & =\sum _{i=1}^{N}\overbrace{( f( x-[ x] +i) -f( x-[ x] +i-1))}^{h( i)} +\sum _{i=N+1}^{[ x]}( f( x-[ x] +i) -f( x-[ x] +i-1)) +\color{red}{( \ )} \end{align*} It follows that

\begin{align*} |\frac{f( x)}{x^{k+1}} -\frac{l}{k+1} | & \leq \sum _{i=1}^{N} |\frac{h( i)}{x^{k+1}} -\frac{l}{[ x]( k+1)} |+\sum _{i=N+1}^{[ x]} |\frac{h( i)}{x^{k+1}} -\frac{l}{[x](k+1)} |+\color{red}{( )x^{-k-1}}\\ & \leq {\textstyle \frac{\overbrace{M}^{\sup _{1\leq i\leq N} h( i)}}{x^{k+1}} +\frac{N|l|}{[ x]( k+1)} +\frac{\epsilon ([ x] -N)}{x} +|\frac{l}{x} -\frac{l}{[ x]( k+1)} |([ x] -N) +\color{red}{\overbrace{\color{red}{\sup _{t\in [ 0,1)}f(t)}}^{M'} x^{-k-1}}}\\ & =\frac{M+M'}{x^{k+1}} +{\textstyle \frac{N|l|}{[ x]( k+1)} +\color{purple}{|\frac{l[ x]}{x} -\frac{l}{( k+1)} |} -N|\frac{l}{x} -\frac{l}{[ x]( k+1)} |+\frac{\epsilon ([ x] -N)}{x}} \tag 1 \end{align*}

In $(1)$, except the purple term every other term can be made arbitrarily small. How do I take care of the purple term so that I can conlude the desired result by limit definition.

Answer 1

First we prove the statement for the case $l=0$, i.e. $ \lim_{n \to \infty}\frac{f(x+1)-f(x)}{x^k} = 0 $.

Given $\epsilon > 0$ there is a $y > a$ such that for all $x \ge y$: $$ \left | \frac{f(x+1)-f(x)}{x^k} \right | < \epsilon \, . $$ Similarly as in If $\lim_{x\to+\infty}[f(x+1)-f(x)]= \ell,$ then $\lim\limits_{x\to+\infty}\frac{f(x)}x=\ell$. we write $$ \frac{f(x)}{x^{k+1}} = \frac{1}{x}\left(\sum_{i=1}^{\lfloor x-y\rfloor}\frac{f(x-i+1)-f(x-i)}{(x-i)^k} \cdot \left( 1-\frac ix\right)^k \right)+\frac{f(x-\lfloor x-y\rfloor)}{x^{k+1}} $$ which implies $$ \left| \frac{f(x)}{x^{k+1}}\right| \le \frac{\lfloor x-y\rfloor}{x} \epsilon + \frac{M}{x^{k+1}} $$ with $M = \sup \{ |f(z)| : y \le z < y+1\}$. It follows that $$ \limsup_{n \to \infty }\left| \frac{f(x)}{x^{k+1}}\right| \le \epsilon \, . $$ This holds for all $\epsilon > 0$, and therefore $$ \lim_{n \to \infty} \frac{f(x)}{x^{k+1}} = 0 \, . $$

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For the general case with $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l$ we set $$ g(x) = f(x) - \frac{x^{k+1}}{k+1} l \, . $$ $g$ is bounded on every finite interval $(a, b)$ and satisfies $$ \frac{g(x+1)-g(x)}{x^k} = \frac{f(x+1)-f(x)}{x^k} - l \frac{(x+1)^{k+1}-x^{k+1}}{(k+1)x^k} \\ = \frac{f(x+1)-f(x)}{x^k} - l \frac{(1+1/x)^{k+1}-1}{(k+1)/x} \to l - l = 0 $$ and then $$ \frac{f(x)}{x^{k+1}} = \frac{g(x)}{x^{k+1}} + \frac{l}{k+1} \to 0 + \frac{l}{k+1} = \frac{l}{k+1} \, . $$

Answer 2

The following is an alternative approach and does not really try to fix the flaw in yours. I am not sure if that can be fixed easily.

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As mentioned in comments to Martin's answer we can prove more generally the analogue of Cesaro-Stolz :

Theorem: Let $f, g$ be functions defined on interval $(a, \infty) $ such that $f$ is bounded on any bounded subinterval of $(a, \infty) $ and let $g$ be strictly increasing with $g(x) \to\infty $ as $x\to\infty $. Then we have $$\lim_{x\to \infty} \frac{f(x+1)-f(x)}{g(x+1)-g(x)}\to L\implies \lim_{x\to\infty} \frac{f(x)} {g(x)} \to L.$$

Let $\epsilon >0$ and then we have a number $y>a$ such that $$L-\epsilon <\frac{f(x+1)-f(x)}{g(x+1)-g(x)}<L+\epsilon $$ and $g(x) >0$ for all $x\geq y$. Since $g$ is strictly increasing the above leads us to $$(L-\epsilon) (g(x+1)-g(x))<f(x+1)-f(x)<(L+\epsilon) (g(x+1)-g(x))\tag{1}$$ Let $m=\lfloor x-y\rfloor$ so that $m$ is a non-negative integer depending on $x$ (if one wants to be explicit one can write $m_x$ instead of $m$) and $$y\leq x-m< y+1$$ Then we have for each value of $i=0,1,2,\dots,m$ $$(L-\epsilon) (g(x+1-i)-g(x-i))<f(x+1-i)-f(x-i)<(L+\epsilon) (g(x+1-i)-g(x-i))$$ This is simply obtained by replacing $x$ with $x-i$ in $(1)$. Adding the above $m+1$ inequalities we get $$(L-\epsilon) (g(x+1)-g(x-m))<f(x+1)-f(x-m)<(L+\epsilon)(g(x+1)-g(x-m))$$ Dividing the above by $g(x+1)$ we get $$(L-\epsilon) \left(1-\frac{g(x-m)}{g(x+1)}\right)<\frac{f(x+1)}{g(x+1)}-\frac{f(x-m)}{g(x+1)}<(L+\epsilon) \left(1-\frac{g(x-m)}{g(x+1)}\right)$$ Taking limits as $x\to\infty $ and noting that $x-m\in[y, y+1]$ and $f, g$ being bounded in $[y, y+1]$ ($g$ is bounded because of monotone nature) we get $$L-\epsilon \leq \liminf_{x\to\infty} \frac{f(x+1)}{g(x+1)}\leq \limsup_{x\to\infty} \frac{f(x+1)}{g(x+1)}\leq L+\epsilon $$ Since $\epsilon >0$ is arbitrary it follows that $f(x) /g(x) \to L$ as $x\to\infty $.

The proof can be adapted to deal with a strictly decreasing $g$ which tends to $-\infty $. Also the argument $x+1$ can be replaced by $x+c$ for any non-zero $c$. And we have a similar result for $x\to-\infty $.

The current question is handled by setting $g(x) =x^{k+1}/(k+1)$.

Answer 3

take $f(x)=ax^{k+1}+bx^{k}+cx^{k-1}+...$ $$degree(f(x))=k+1$$and degree of $$f(x+1)-f(x)=a(x+1)^{k+1}+b(x+1)^k+c(x+1)^{k-1}+...\\-(ax^{k+1}+bx^{k}+cx^{k-1}+...)\\=(a-a)\not{x^{k+1}}+a\times{^{k+1}}C_1 x^{k}+...\\=a(k+1)x^k+...$$so $$\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l\\ \lim_{x\to \infty} \frac{a(k+1)x^k+...}{x^k}=l\\a(k+1)=l\\a=\frac{l}{k+1}$$so $$\lim_{x\to \infty} \frac{f(x)}{x^{k+1}}=\lim_{x\to \infty} \frac{ax^{k+1}+bx^{k}+cx^{k-1}+...}{x^{k+1}}=a=\frac{l}{k+1}\\$$