$underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ Implies $underset{xrightarrowinfty}{f'(x)}=0$ [duplicate]
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This question already has an answer here:
Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist
6 answers
Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuously differentiable function such that $underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ and suppose $underset{xrightarrowinfty}{f'(x)}$ exists. Then Prove that $underset{xrightarrowinfty}{f'(x)}=0$
I can see that if we apply L'hoptal's theorem directly to $frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $underset{xrightarrowinfty}{f(x)}$
On the similar problem: found here, they have given the existence of $lim_{xrightarrowinfty} f(x)$. But in this particular problem they haven't
real-analysis calculus analysis
asked Jan 2 at 21:30
DD90DD90
2648
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marked as duplicate by Math Lover, Cesareo, José Carlos Santos calculus
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Jan 3 at 1:17
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 |
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This question already has an answer here:
Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist
6 answers
Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuously differentiable function such that $underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ and suppose $underset{xrightarrowinfty}{f'(x)}$ exists. Then Prove that $underset{xrightarrowinfty}{f'(x)}=0$
I can see that if we apply L'hoptal's theorem directly to $frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $underset{xrightarrowinfty}{f(x)}$
On the similar problem: found here, they have given the existence of $lim_{xrightarrowinfty} f(x)$. But in this particular problem they haven't
real-analysis calculus analysis
asked Jan 2 at 21:30
DD90DD90
2648
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marked as duplicate by Math Lover, Cesareo, José Carlos Santos calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Jan 3 at 1:17
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.
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It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
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– Martin R
Jan 2 at 21:35
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Thanks and I edited
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– DD90
Jan 2 at 21:37
add a comment |
$begingroup$
This question already has an answer here:
Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist
6 answers
Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuously differentiable function such that $underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ and suppose $underset{xrightarrowinfty}{f'(x)}$ exists. Then Prove that $underset{xrightarrowinfty}{f'(x)}=0$
I can see that if we apply L'hoptal's theorem directly to $frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $underset{xrightarrowinfty}{f(x)}$
On the similar problem: found here, they have given the existence of $lim_{xrightarrowinfty} f(x)$. But in this particular problem they haven't
real-analysis calculus analysis
asked Jan 2 at 21:30
DD90DD90
2648
$endgroup$
This question already has an answer here:
Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist
6 answers
Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuously differentiable function such that $underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ and suppose $underset{xrightarrowinfty}{f'(x)}$ exists. Then Prove that $underset{xrightarrowinfty}{f'(x)}=0$
I can see that if we apply L'hoptal's theorem directly to $frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $underset{xrightarrowinfty}{f(x)}$
On the similar problem: found here, they have given the existence of $lim_{xrightarrowinfty} f(x)$. But in this particular problem they haven't
This question already has an answer here:
Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist
6 answers
real-analysis calculus analysis
real-analysis calculus analysis
asked Jan 2 at 21:30
DD90DD90
2648
asked Jan 2 at 21:30
DD90DD90
2648
edited Jan 2 at 21:52
DD90
asked Jan 2 at 21:30
DD90DD90
2648
asked Jan 2 at 21:30
DD90DD90
2648
asked Jan 2 at 21:30
DD90DD90
2648
2648
marked as duplicate by Math Lover, Cesareo, José Carlos Santos calculus
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Jan 3 at 1:17
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 Math Lover, Cesareo, José Carlos Santos calculus
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Jan 3 at 1:17
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.
$begingroup$
It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
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– Martin R
Jan 2 at 21:35
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Thanks and I edited
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– DD90
Jan 2 at 21:37
add a comment |
$begingroup$
It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
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– Martin R
Jan 2 at 21:35
$begingroup$
Thanks and I edited
$endgroup$
– DD90
Jan 2 at 21:37
$begingroup$
It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
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– Martin R
Jan 2 at 21:35
$begingroup$
It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
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– Martin R
Jan 2 at 21:35
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Thanks and I edited
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– DD90
Jan 2 at 21:37
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Thanks and I edited
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– DD90
Jan 2 at 21:37
add a comment |
2 Answers
2
active
oldest
votes
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As stated the result is wrong. Take $f(x)=sin x^2$.
answered Jan 2 at 21:34
mathcounterexamples.netmathcounterexamples.net
27k22158
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$begingroup$
Sorry I edited.
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– DD90
Jan 2 at 21:52
add a comment |
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Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.
answered Jan 2 at 23:51
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As stated the result is wrong. Take $f(x)=sin x^2$.
answered Jan 2 at 21:34
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
Sorry I edited.
$endgroup$
– DD90
Jan 2 at 21:52
add a comment |
$begingroup$
As stated the result is wrong. Take $f(x)=sin x^2$.
answered Jan 2 at 21:34
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
Sorry I edited.
$endgroup$
– DD90
Jan 2 at 21:52
add a comment |
$begingroup$
As stated the result is wrong. Take $f(x)=sin x^2$.
answered Jan 2 at 21:34
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
As stated the result is wrong. Take $f(x)=sin x^2$.
answered Jan 2 at 21:34
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 2 at 21:34
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 2 at 21:34
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 2 at 21:34
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
$begingroup$
Sorry I edited.
$endgroup$
– DD90
Jan 2 at 21:52
add a comment |
$begingroup$
Sorry I edited.
$endgroup$
– DD90
Jan 2 at 21:52
$begingroup$
Sorry I edited.
$endgroup$
– DD90
Jan 2 at 21:52
$begingroup$
Sorry I edited.
$endgroup$
– DD90
Jan 2 at 21:52
add a comment |
$begingroup$
Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.
answered Jan 2 at 23:51
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
add a comment |
$begingroup$
Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.
answered Jan 2 at 23:51
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
add a comment |
$begingroup$
Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.
answered Jan 2 at 23:51
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.
answered Jan 2 at 23:51
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Jan 2 at 23:51
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Jan 2 at 23:51
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Jan 2 at 23:51
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
71.4k53170
add a comment |
add a comment |
$begingroup$
It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
$endgroup$
– Martin R
Jan 2 at 21:35
$begingroup$
Thanks and I edited
$endgroup$
– DD90
Jan 2 at 21:37