Prove that $lim_{x to infty} f(x)/x=lim_{x to infty}[f(x+1)-f(x)]$ [duplicate]
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This question already has an answer here:
Prove that $lim_{x to +infty} frac{f(x)}{x} = L$ if $lim_{x to +infty} [f(x+1) - f(x)] = L space$
2 answers
Let $f:(a,infty)rightarrow mathbb{R}$ be a function such that $f$ is bounded in any finite interval $(a,b]$. Prove that $lim_{x to infty} f(x)/x=lim_{x to infty}[f(x+1)-f(x)]$, provided that the right limit exists. Thanks a lot.
real-analysis calculus limits
edited Dec 20 '18 at 20:21
JavaMan
11.2k12855
asked Dec 20 '18 at 19:10
S Ali MousaviS Ali Mousavi
456411
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marked as duplicate by grand_chat, Community♦ Dec 20 '18 at 20:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Prove that $lim_{x to +infty} frac{f(x)}{x} = L$ if $lim_{x to +infty} [f(x+1) - f(x)] = L space$
2 answers
Let $f:(a,infty)rightarrow mathbb{R}$ be a function such that $f$ is bounded in any finite interval $(a,b]$. Prove that $lim_{x to infty} f(x)/x=lim_{x to infty}[f(x+1)-f(x)]$, provided that the right limit exists. Thanks a lot.
real-analysis calculus limits
edited Dec 20 '18 at 20:21
JavaMan
11.2k12855
asked Dec 20 '18 at 19:10
S Ali MousaviS Ali Mousavi
456411
$endgroup$
marked as duplicate by grand_chat, Community♦ Dec 20 '18 at 20:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Maybe you are missing a hypothesis. Let $f(x)=sin{x} +1$. The left hand side is zero and the right hand side is not defined. On the other hand if you add the hypothesis that $lim_{xrightarrow infty}f’(x)$ exists, it’s true.
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– Charlie Frohman
Dec 20 '18 at 19:20
2
$begingroup$
That requires differentiability, and the question doesn't even require continuity. Might there be a slightly better approach, perhaps $f$ is strictly monotonic? @CharlieFrohman I suppose then, we'd have the bounded part by default.
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– Thomas Andrews
Dec 20 '18 at 19:24
2
$begingroup$
Now, what have you tried?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:30
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Thanks. I edit my question. My attempt is to write $f(x+1)-f(x)=L+e(x)$ such that $lim_{x to infty} e(x)=0$. But i could not prove it.
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– S Ali Mousavi
Dec 20 '18 at 19:32
add a comment |
$begingroup$
This question already has an answer here:
Prove that $lim_{x to +infty} frac{f(x)}{x} = L$ if $lim_{x to +infty} [f(x+1) - f(x)] = L space$
2 answers
Let $f:(a,infty)rightarrow mathbb{R}$ be a function such that $f$ is bounded in any finite interval $(a,b]$. Prove that $lim_{x to infty} f(x)/x=lim_{x to infty}[f(x+1)-f(x)]$, provided that the right limit exists. Thanks a lot.
real-analysis calculus limits
edited Dec 20 '18 at 20:21
JavaMan
11.2k12855
asked Dec 20 '18 at 19:10
S Ali MousaviS Ali Mousavi
456411
$endgroup$
This question already has an answer here:
Prove that $lim_{x to +infty} frac{f(x)}{x} = L$ if $lim_{x to +infty} [f(x+1) - f(x)] = L space$
2 answers
Let $f:(a,infty)rightarrow mathbb{R}$ be a function such that $f$ is bounded in any finite interval $(a,b]$. Prove that $lim_{x to infty} f(x)/x=lim_{x to infty}[f(x+1)-f(x)]$, provided that the right limit exists. Thanks a lot.
This question already has an answer here:
Prove that $lim_{x to +infty} frac{f(x)}{x} = L$ if $lim_{x to +infty} [f(x+1) - f(x)] = L space$
2 answers
real-analysis calculus limits
real-analysis calculus limits
edited Dec 20 '18 at 20:21
JavaMan
11.2k12855
asked Dec 20 '18 at 19:10
S Ali MousaviS Ali Mousavi
456411
edited Dec 20 '18 at 20:21
JavaMan
11.2k12855
asked Dec 20 '18 at 19:10
S Ali MousaviS Ali Mousavi
456411
edited Dec 20 '18 at 20:21
JavaMan
11.2k12855
edited Dec 20 '18 at 20:21
JavaMan
11.2k12855
edited Dec 20 '18 at 20:21
JavaMan
11.2k12855
11.2k12855
asked Dec 20 '18 at 19:10
S Ali MousaviS Ali Mousavi
456411
asked Dec 20 '18 at 19:10
S Ali MousaviS Ali Mousavi
456411
asked Dec 20 '18 at 19:10
S Ali MousaviS Ali Mousavi
456411
456411
marked as duplicate by grand_chat, Community♦ Dec 20 '18 at 20:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by grand_chat, Community♦ Dec 20 '18 at 20:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Maybe you are missing a hypothesis. Let $f(x)=sin{x} +1$. The left hand side is zero and the right hand side is not defined. On the other hand if you add the hypothesis that $lim_{xrightarrow infty}f’(x)$ exists, it’s true.
$endgroup$
– Charlie Frohman
Dec 20 '18 at 19:20
2
$begingroup$
That requires differentiability, and the question doesn't even require continuity. Might there be a slightly better approach, perhaps $f$ is strictly monotonic? @CharlieFrohman I suppose then, we'd have the bounded part by default.
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:24
2
$begingroup$
Now, what have you tried?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:30
$begingroup$
Thanks. I edit my question. My attempt is to write $f(x+1)-f(x)=L+e(x)$ such that $lim_{x to infty} e(x)=0$. But i could not prove it.
$endgroup$
– S Ali Mousavi
Dec 20 '18 at 19:32
add a comment |
2
$begingroup$
Maybe you are missing a hypothesis. Let $f(x)=sin{x} +1$. The left hand side is zero and the right hand side is not defined. On the other hand if you add the hypothesis that $lim_{xrightarrow infty}f’(x)$ exists, it’s true.
$endgroup$
– Charlie Frohman
Dec 20 '18 at 19:20
2
$begingroup$
That requires differentiability, and the question doesn't even require continuity. Might there be a slightly better approach, perhaps $f$ is strictly monotonic? @CharlieFrohman I suppose then, we'd have the bounded part by default.
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:24
2
$begingroup$
Now, what have you tried?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:30
$begingroup$
Thanks. I edit my question. My attempt is to write $f(x+1)-f(x)=L+e(x)$ such that $lim_{x to infty} e(x)=0$. But i could not prove it.
$endgroup$
– S Ali Mousavi
Dec 20 '18 at 19:32
2
2
$begingroup$
Maybe you are missing a hypothesis. Let $f(x)=sin{x} +1$. The left hand side is zero and the right hand side is not defined. On the other hand if you add the hypothesis that $lim_{xrightarrow infty}f’(x)$ exists, it’s true.
$endgroup$
– Charlie Frohman
Dec 20 '18 at 19:20
$begingroup$
Maybe you are missing a hypothesis. Let $f(x)=sin{x} +1$. The left hand side is zero and the right hand side is not defined. On the other hand if you add the hypothesis that $lim_{xrightarrow infty}f’(x)$ exists, it’s true.
$endgroup$
– Charlie Frohman
Dec 20 '18 at 19:20
2
2
$begingroup$
That requires differentiability, and the question doesn't even require continuity. Might there be a slightly better approach, perhaps $f$ is strictly monotonic? @CharlieFrohman I suppose then, we'd have the bounded part by default.
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:24
$begingroup$
That requires differentiability, and the question doesn't even require continuity. Might there be a slightly better approach, perhaps $f$ is strictly monotonic? @CharlieFrohman I suppose then, we'd have the bounded part by default.
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:24
2
2
$begingroup$
Now, what have you tried?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:30
$begingroup$
Now, what have you tried?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:30
$begingroup$
Thanks. I edit my question. My attempt is to write $f(x+1)-f(x)=L+e(x)$ such that $lim_{x to infty} e(x)=0$. But i could not prove it.
$endgroup$
– S Ali Mousavi
Dec 20 '18 at 19:32
$begingroup$
Thanks. I edit my question. My attempt is to write $f(x+1)-f(x)=L+e(x)$ such that $lim_{x to infty} e(x)=0$. But i could not prove it.
$endgroup$
– S Ali Mousavi
Dec 20 '18 at 19:32
add a comment |
1 Answer
1
active
oldest
votes
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Let $c=lim f(x+y)-f(x)$. Given $epsilon>0$, there is $M$ such that $|f(x+y)-f(x)-c|<epsilon$ for $x>M$. Also, $|f(x)| < L$ for $xle M+1$. Then for $x>M$ and with $n=lfloor x-Mrfloor$,
$$-L+n(c-epsilon)<f(x-n)+n(c-epsilon)le f(x)le f(x-n)+n(c+epsilon)<L+n(c+epsilon) $$
and so
$$ frac {L-M(|c|+epsilon)}x+c-epsilon<frac{f(x)}x<frac {L+M(|c|+epsilon)}x+c+epsilon$$
which implies
$$ c-epsilonle liminf frac{f(x)}x, quadlimsup frac{f(x)}xle c+epsilon.$$
As $epsilon$ was arbitrary, $$ limfrac{f(x)}x=c.$$
answered Dec 20 '18 at 19:45
Hagen von EitzenHagen von Eitzen
2943
$endgroup$
$begingroup$
How do you know $c$ exists? It only necessarily exists for $y$ an integer, right?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:49
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $c=lim f(x+y)-f(x)$. Given $epsilon>0$, there is $M$ such that $|f(x+y)-f(x)-c|<epsilon$ for $x>M$. Also, $|f(x)| < L$ for $xle M+1$. Then for $x>M$ and with $n=lfloor x-Mrfloor$,
$$-L+n(c-epsilon)<f(x-n)+n(c-epsilon)le f(x)le f(x-n)+n(c+epsilon)<L+n(c+epsilon) $$
and so
$$ frac {L-M(|c|+epsilon)}x+c-epsilon<frac{f(x)}x<frac {L+M(|c|+epsilon)}x+c+epsilon$$
which implies
$$ c-epsilonle liminf frac{f(x)}x, quadlimsup frac{f(x)}xle c+epsilon.$$
As $epsilon$ was arbitrary, $$ limfrac{f(x)}x=c.$$
answered Dec 20 '18 at 19:45
Hagen von EitzenHagen von Eitzen
2943
$endgroup$
$begingroup$
How do you know $c$ exists? It only necessarily exists for $y$ an integer, right?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:49
add a comment |
$begingroup$
Let $c=lim f(x+y)-f(x)$. Given $epsilon>0$, there is $M$ such that $|f(x+y)-f(x)-c|<epsilon$ for $x>M$. Also, $|f(x)| < L$ for $xle M+1$. Then for $x>M$ and with $n=lfloor x-Mrfloor$,
$$-L+n(c-epsilon)<f(x-n)+n(c-epsilon)le f(x)le f(x-n)+n(c+epsilon)<L+n(c+epsilon) $$
and so
$$ frac {L-M(|c|+epsilon)}x+c-epsilon<frac{f(x)}x<frac {L+M(|c|+epsilon)}x+c+epsilon$$
which implies
$$ c-epsilonle liminf frac{f(x)}x, quadlimsup frac{f(x)}xle c+epsilon.$$
As $epsilon$ was arbitrary, $$ limfrac{f(x)}x=c.$$
answered Dec 20 '18 at 19:45
Hagen von EitzenHagen von Eitzen
2943
$endgroup$
$begingroup$
How do you know $c$ exists? It only necessarily exists for $y$ an integer, right?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:49
add a comment |
$begingroup$
Let $c=lim f(x+y)-f(x)$. Given $epsilon>0$, there is $M$ such that $|f(x+y)-f(x)-c|<epsilon$ for $x>M$. Also, $|f(x)| < L$ for $xle M+1$. Then for $x>M$ and with $n=lfloor x-Mrfloor$,
$$-L+n(c-epsilon)<f(x-n)+n(c-epsilon)le f(x)le f(x-n)+n(c+epsilon)<L+n(c+epsilon) $$
and so
$$ frac {L-M(|c|+epsilon)}x+c-epsilon<frac{f(x)}x<frac {L+M(|c|+epsilon)}x+c+epsilon$$
which implies
$$ c-epsilonle liminf frac{f(x)}x, quadlimsup frac{f(x)}xle c+epsilon.$$
As $epsilon$ was arbitrary, $$ limfrac{f(x)}x=c.$$
answered Dec 20 '18 at 19:45
Hagen von EitzenHagen von Eitzen
2943
$endgroup$
Let $c=lim f(x+y)-f(x)$. Given $epsilon>0$, there is $M$ such that $|f(x+y)-f(x)-c|<epsilon$ for $x>M$. Also, $|f(x)| < L$ for $xle M+1$. Then for $x>M$ and with $n=lfloor x-Mrfloor$,
$$-L+n(c-epsilon)<f(x-n)+n(c-epsilon)le f(x)le f(x-n)+n(c+epsilon)<L+n(c+epsilon) $$
and so
$$ frac {L-M(|c|+epsilon)}x+c-epsilon<frac{f(x)}x<frac {L+M(|c|+epsilon)}x+c+epsilon$$
which implies
$$ c-epsilonle liminf frac{f(x)}x, quadlimsup frac{f(x)}xle c+epsilon.$$
As $epsilon$ was arbitrary, $$ limfrac{f(x)}x=c.$$
answered Dec 20 '18 at 19:45
Hagen von EitzenHagen von Eitzen
2943
answered Dec 20 '18 at 19:45
Hagen von EitzenHagen von Eitzen
2943
answered Dec 20 '18 at 19:45
Hagen von EitzenHagen von Eitzen
2943
answered Dec 20 '18 at 19:45
Hagen von EitzenHagen von Eitzen
2943
2943
$begingroup$
How do you know $c$ exists? It only necessarily exists for $y$ an integer, right?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:49
add a comment |
$begingroup$
How do you know $c$ exists? It only necessarily exists for $y$ an integer, right?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:49
$begingroup$
How do you know $c$ exists? It only necessarily exists for $y$ an integer, right?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:49
$begingroup$
How do you know $c$ exists? It only necessarily exists for $y$ an integer, right?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:49
add a comment |
2
$begingroup$
Maybe you are missing a hypothesis. Let $f(x)=sin{x} +1$. The left hand side is zero and the right hand side is not defined. On the other hand if you add the hypothesis that $lim_{xrightarrow infty}f’(x)$ exists, it’s true.
$endgroup$
– Charlie Frohman
Dec 20 '18 at 19:20
2
$begingroup$
That requires differentiability, and the question doesn't even require continuity. Might there be a slightly better approach, perhaps $f$ is strictly monotonic? @CharlieFrohman I suppose then, we'd have the bounded part by default.
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:24
2
$begingroup$
Now, what have you tried?
$endgroup$
– Thomas Andrews
Dec 20 '18 at 19:30
$begingroup$
Thanks. I edit my question. My attempt is to write $f(x+1)-f(x)=L+e(x)$ such that $lim_{x to infty} e(x)=0$. But i could not prove it.
$endgroup$
– S Ali Mousavi
Dec 20 '18 at 19:32