Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?
$begingroup$
Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?
I think that this is true. Because $f$ u.c. implies $f$ has bounded derivative. Can someone help me confirm this hypothesis?
Thanks
real-analysis limits
$endgroup$
add a comment |
$begingroup$
Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?
I think that this is true. Because $f$ u.c. implies $f$ has bounded derivative. Can someone help me confirm this hypothesis?
Thanks
real-analysis limits
$endgroup$
3
$begingroup$
$sin x{{{}}}$?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 17:44
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oh that is a counterexample. oops thanks
$endgroup$
– stackofhay42
Dec 14 '18 at 17:45
$begingroup$
wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:46
$begingroup$
Yes that is true stack.
$endgroup$
– zhw.
Dec 14 '18 at 17:48
add a comment |
$begingroup$
Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?
I think that this is true. Because $f$ u.c. implies $f$ has bounded derivative. Can someone help me confirm this hypothesis?
Thanks
real-analysis limits
$endgroup$
Does $f$ uniformly continuous on $[0, infty)$ imply that $lim_{xtoinfty} f(x)$ exists?
I think that this is true. Because $f$ u.c. implies $f$ has bounded derivative. Can someone help me confirm this hypothesis?
Thanks
real-analysis limits
real-analysis limits
asked Dec 14 '18 at 17:42
stackofhay42stackofhay42
2177
2177
3
$begingroup$
$sin x{{{}}}$?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 17:44
$begingroup$
oh that is a counterexample. oops thanks
$endgroup$
– stackofhay42
Dec 14 '18 at 17:45
$begingroup$
wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:46
$begingroup$
Yes that is true stack.
$endgroup$
– zhw.
Dec 14 '18 at 17:48
add a comment |
3
$begingroup$
$sin x{{{}}}$?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 17:44
$begingroup$
oh that is a counterexample. oops thanks
$endgroup$
– stackofhay42
Dec 14 '18 at 17:45
$begingroup$
wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:46
$begingroup$
Yes that is true stack.
$endgroup$
– zhw.
Dec 14 '18 at 17:48
3
3
$begingroup$
$sin x{{{}}}$?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 17:44
$begingroup$
$sin x{{{}}}$?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 17:44
$begingroup$
oh that is a counterexample. oops thanks
$endgroup$
– stackofhay42
Dec 14 '18 at 17:45
$begingroup$
oh that is a counterexample. oops thanks
$endgroup$
– stackofhay42
Dec 14 '18 at 17:45
$begingroup$
wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:46
$begingroup$
wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:46
$begingroup$
Yes that is true stack.
$endgroup$
– zhw.
Dec 14 '18 at 17:48
$begingroup$
Yes that is true stack.
$endgroup$
– zhw.
Dec 14 '18 at 17:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No—consider the function $sin(x)$.
It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.
$endgroup$
$begingroup$
so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:47
$begingroup$
@stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:51
add a comment |
$begingroup$
The function $$f : xmapsto sqrt{x}$$
is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No—consider the function $sin(x)$.
It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.
$endgroup$
$begingroup$
so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:47
$begingroup$
@stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:51
add a comment |
$begingroup$
No—consider the function $sin(x)$.
It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.
$endgroup$
$begingroup$
so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:47
$begingroup$
@stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:51
add a comment |
$begingroup$
No—consider the function $sin(x)$.
It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.
$endgroup$
No—consider the function $sin(x)$.
It does not converge because it oscillates at infinity, but it clearly is uniformly continuous, being differentiable everywhere and having bounded derivative.
answered Dec 14 '18 at 17:46
AlexAlex
1773
1773
$begingroup$
so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:47
$begingroup$
@stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:51
add a comment |
$begingroup$
so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:47
$begingroup$
@stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:51
$begingroup$
so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:47
$begingroup$
so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:47
$begingroup$
@stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:51
$begingroup$
@stackofhay42 Sorry, this is not related to your question directly but I've seen your trend in the past of never accepting an answer to any of your question. If an answer is complete and clear your doubt you should consider accepting it so that your question will be marked as answered.
$endgroup$
– BigbearZzz
Dec 14 '18 at 17:51
add a comment |
$begingroup$
The function $$f : xmapsto sqrt{x}$$
is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$
$endgroup$
add a comment |
$begingroup$
The function $$f : xmapsto sqrt{x}$$
is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$
$endgroup$
add a comment |
$begingroup$
The function $$f : xmapsto sqrt{x}$$
is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$
$endgroup$
The function $$f : xmapsto sqrt{x}$$
is uniformly continuous at $[0,+infty)$ but $$lim_{xto+infty}f(x)=+infty$$
answered Dec 14 '18 at 18:30
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
add a comment |
add a comment |
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3
$begingroup$
$sin x{{{}}}$?
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 17:44
$begingroup$
oh that is a counterexample. oops thanks
$endgroup$
– stackofhay42
Dec 14 '18 at 17:45
$begingroup$
wait so then if $f$ is continuous on $[0, infty)$ AND the $lim_{xtoinfty} f(x) = a$ for some $a in mathbb{R}$, then is it true that $f$ is uniformly continuous? this must be true then, right?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:46
$begingroup$
Yes that is true stack.
$endgroup$
– zhw.
Dec 14 '18 at 17:48