How can $limsup_{x to x_0} f(x) = f(x_0)$ for $f$ discontinuous at $x_0$?












0












$begingroup$


My textbook says
$$limsup_{x to x_0} f(x) = max{f(x_0), lim_{hto 0^+} f( x_0 + h), lim_{hto 0^-} f( x_0 + h)}$$



Assuming $f(x_0)$ is distinct from the latter two values, how can $limsup f(x)$ as $x$ approaches $x_0$ equal neither $limsup f(x)$ approaching $x_0$ from the left or from the right? Doesn't that violate the definition of a limit, since $f(x_0)$ isn't "approached?"



For clarifcation, assuming wlog $f(x_0) < f(x)$, $sup f(x) = max{(f(x_0), f(x))}$. The limit in question is the value approached as $x$ approaches $x_0$.



Answer: $limsup f(x)$ has to include $f(x_0)$ in order for the statement "if $limsup f(x_0) = liminf f(x_0)$, then $f$ is continuous at $x_0$" to be always true.






real-analysis limsup-and-liminf







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What does $lim_{xto x_0}sup f(x)$ even mean? What is the $sup$ over if this is a limit?
    $endgroup$
    – Don Thousand
    Dec 5 '18 at 21:08












  • $begingroup$
    Why is this so hard to happen? Imagine f is just constant on a sequence tending to $x_0$, the constant being less than $f(x_0)$ and is equal to the identity function otherwise.
    $endgroup$
    – Sorin Tirc
    Dec 5 '18 at 21:11










  • $begingroup$
    Please use limsup and even limsuplimits_{xto x_0}.
    $endgroup$
    – Did
    Dec 5 '18 at 21:26










  • $begingroup$
    @Masacroso Sorry but your edit is bad. More care, please.
    $endgroup$
    – Did
    Dec 5 '18 at 21:26










  • $begingroup$
    @Did I dont think it is bad, but if you think that my edit is bad then edit yourself according to your criteria
    $endgroup$
    – Masacroso
    Dec 5 '18 at 21:28


















0












$begingroup$


My textbook says
$$limsup_{x to x_0} f(x) = max{f(x_0), lim_{hto 0^+} f( x_0 + h), lim_{hto 0^-} f( x_0 + h)}$$



Assuming $f(x_0)$ is distinct from the latter two values, how can $limsup f(x)$ as $x$ approaches $x_0$ equal neither $limsup f(x)$ approaching $x_0$ from the left or from the right? Doesn't that violate the definition of a limit, since $f(x_0)$ isn't "approached?"



For clarifcation, assuming wlog $f(x_0) < f(x)$, $sup f(x) = max{(f(x_0), f(x))}$. The limit in question is the value approached as $x$ approaches $x_0$.



Answer: $limsup f(x)$ has to include $f(x_0)$ in order for the statement "if $limsup f(x_0) = liminf f(x_0)$, then $f$ is continuous at $x_0$" to be always true.






real-analysis limsup-and-liminf







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What does $lim_{xto x_0}sup f(x)$ even mean? What is the $sup$ over if this is a limit?
    $endgroup$
    – Don Thousand
    Dec 5 '18 at 21:08












  • $begingroup$
    Why is this so hard to happen? Imagine f is just constant on a sequence tending to $x_0$, the constant being less than $f(x_0)$ and is equal to the identity function otherwise.
    $endgroup$
    – Sorin Tirc
    Dec 5 '18 at 21:11










  • $begingroup$
    Please use limsup and even limsuplimits_{xto x_0}.
    $endgroup$
    – Did
    Dec 5 '18 at 21:26










  • $begingroup$
    @Masacroso Sorry but your edit is bad. More care, please.
    $endgroup$
    – Did
    Dec 5 '18 at 21:26










  • $begingroup$
    @Did I dont think it is bad, but if you think that my edit is bad then edit yourself according to your criteria
    $endgroup$
    – Masacroso
    Dec 5 '18 at 21:28
















0












0








0





$begingroup$


My textbook says
$$limsup_{x to x_0} f(x) = max{f(x_0), lim_{hto 0^+} f( x_0 + h), lim_{hto 0^-} f( x_0 + h)}$$



Assuming $f(x_0)$ is distinct from the latter two values, how can $limsup f(x)$ as $x$ approaches $x_0$ equal neither $limsup f(x)$ approaching $x_0$ from the left or from the right? Doesn't that violate the definition of a limit, since $f(x_0)$ isn't "approached?"



For clarifcation, assuming wlog $f(x_0) < f(x)$, $sup f(x) = max{(f(x_0), f(x))}$. The limit in question is the value approached as $x$ approaches $x_0$.



Answer: $limsup f(x)$ has to include $f(x_0)$ in order for the statement "if $limsup f(x_0) = liminf f(x_0)$, then $f$ is continuous at $x_0$" to be always true.






real-analysis limsup-and-liminf







share|cite|improve this question











$endgroup$




My textbook says
$$limsup_{x to x_0} f(x) = max{f(x_0), lim_{hto 0^+} f( x_0 + h), lim_{hto 0^-} f( x_0 + h)}$$



Assuming $f(x_0)$ is distinct from the latter two values, how can $limsup f(x)$ as $x$ approaches $x_0$ equal neither $limsup f(x)$ approaching $x_0$ from the left or from the right? Doesn't that violate the definition of a limit, since $f(x_0)$ isn't "approached?"



For clarifcation, assuming wlog $f(x_0) < f(x)$, $sup f(x) = max{(f(x_0), f(x))}$. The limit in question is the value approached as $x$ approaches $x_0$.



Answer: $limsup f(x)$ has to include $f(x_0)$ in order for the statement "if $limsup f(x_0) = liminf f(x_0)$, then $f$ is continuous at $x_0$" to be always true.






real-analysis limsup-and-liminf




real-analysis limsup-and-liminf






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 22:18







hiroshin

















asked Dec 5 '18 at 21:06









hiroshinhiroshin

666




666








  • 2




    $begingroup$
    What does $lim_{xto x_0}sup f(x)$ even mean? What is the $sup$ over if this is a limit?
    $endgroup$
    – Don Thousand
    Dec 5 '18 at 21:08












  • $begingroup$
    Why is this so hard to happen? Imagine f is just constant on a sequence tending to $x_0$, the constant being less than $f(x_0)$ and is equal to the identity function otherwise.
    $endgroup$
    – Sorin Tirc
    Dec 5 '18 at 21:11










  • $begingroup$
    Please use limsup and even limsuplimits_{xto x_0}.
    $endgroup$
    – Did
    Dec 5 '18 at 21:26










  • $begingroup$
    @Masacroso Sorry but your edit is bad. More care, please.
    $endgroup$
    – Did
    Dec 5 '18 at 21:26










  • $begingroup$
    @Did I dont think it is bad, but if you think that my edit is bad then edit yourself according to your criteria
    $endgroup$
    – Masacroso
    Dec 5 '18 at 21:28
















  • 2




    $begingroup$
    What does $lim_{xto x_0}sup f(x)$ even mean? What is the $sup$ over if this is a limit?
    $endgroup$
    – Don Thousand
    Dec 5 '18 at 21:08












  • $begingroup$
    Why is this so hard to happen? Imagine f is just constant on a sequence tending to $x_0$, the constant being less than $f(x_0)$ and is equal to the identity function otherwise.
    $endgroup$
    – Sorin Tirc
    Dec 5 '18 at 21:11










  • $begingroup$
    Please use limsup and even limsuplimits_{xto x_0}.
    $endgroup$
    – Did
    Dec 5 '18 at 21:26










  • $begingroup$
    @Masacroso Sorry but your edit is bad. More care, please.
    $endgroup$
    – Did
    Dec 5 '18 at 21:26










  • $begingroup$
    @Did I dont think it is bad, but if you think that my edit is bad then edit yourself according to your criteria
    $endgroup$
    – Masacroso
    Dec 5 '18 at 21:28










2




2




$begingroup$
What does $lim_{xto x_0}sup f(x)$ even mean? What is the $sup$ over if this is a limit?
$endgroup$
– Don Thousand
Dec 5 '18 at 21:08






$begingroup$
What does $lim_{xto x_0}sup f(x)$ even mean? What is the $sup$ over if this is a limit?
$endgroup$
– Don Thousand
Dec 5 '18 at 21:08














$begingroup$
Why is this so hard to happen? Imagine f is just constant on a sequence tending to $x_0$, the constant being less than $f(x_0)$ and is equal to the identity function otherwise.
$endgroup$
– Sorin Tirc
Dec 5 '18 at 21:11




$begingroup$
Why is this so hard to happen? Imagine f is just constant on a sequence tending to $x_0$, the constant being less than $f(x_0)$ and is equal to the identity function otherwise.
$endgroup$
– Sorin Tirc
Dec 5 '18 at 21:11












$begingroup$
Please use limsup and even limsuplimits_{xto x_0}.
$endgroup$
– Did
Dec 5 '18 at 21:26




$begingroup$
Please use limsup and even limsuplimits_{xto x_0}.
$endgroup$
– Did
Dec 5 '18 at 21:26












$begingroup$
@Masacroso Sorry but your edit is bad. More care, please.
$endgroup$
– Did
Dec 5 '18 at 21:26




$begingroup$
@Masacroso Sorry but your edit is bad. More care, please.
$endgroup$
– Did
Dec 5 '18 at 21:26












$begingroup$
@Did I dont think it is bad, but if you think that my edit is bad then edit yourself according to your criteria
$endgroup$
– Masacroso
Dec 5 '18 at 21:28






$begingroup$
@Did I dont think it is bad, but if you think that my edit is bad then edit yourself according to your criteria
$endgroup$
– Masacroso
Dec 5 '18 at 21:28












1 Answer
1






active

oldest

votes


















1












$begingroup$

Try $$f(x) = begin{cases} sin(1/x) &, x neq 0 \ 2 &, x = 0 end{cases} text{.} $$ What should $limsup_{x rightarrow 0} f(x)$ be?



Or to put this another way, why do you think $x_0$ cannot be a point in the sequence implicitly encoded by $x rightarrow x_0$? We have to go to some pains to dodge this point in the one-sided limits. (We don't take sequences of points, we take sequences of offsets, $h$ all required to be of the same sign, hence nonzero.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The lim sup on each side is 1, so it would make sense for the "general" lim sup to be the greater of the two, which is 1. My book claims the non-sided lim sup should also take $f(0)$ into account, so instead of 1 it's 2. I don't understand why this is preferred.
    $endgroup$
    – hiroshin
    Dec 5 '18 at 21:59










  • $begingroup$
    @hiroshin : How do you feel about the sequence of inputs to $f$: $(x_0, x_0, x_0, dots, x_0, dots)$? Is this a sequence we should use in our lim sup? What value does it put in the set whose supremum we are taking?
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 23:52













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Try $$f(x) = begin{cases} sin(1/x) &, x neq 0 \ 2 &, x = 0 end{cases} text{.} $$ What should $limsup_{x rightarrow 0} f(x)$ be?



Or to put this another way, why do you think $x_0$ cannot be a point in the sequence implicitly encoded by $x rightarrow x_0$? We have to go to some pains to dodge this point in the one-sided limits. (We don't take sequences of points, we take sequences of offsets, $h$ all required to be of the same sign, hence nonzero.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The lim sup on each side is 1, so it would make sense for the "general" lim sup to be the greater of the two, which is 1. My book claims the non-sided lim sup should also take $f(0)$ into account, so instead of 1 it's 2. I don't understand why this is preferred.
    $endgroup$
    – hiroshin
    Dec 5 '18 at 21:59










  • $begingroup$
    @hiroshin : How do you feel about the sequence of inputs to $f$: $(x_0, x_0, x_0, dots, x_0, dots)$? Is this a sequence we should use in our lim sup? What value does it put in the set whose supremum we are taking?
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 23:52


















1












$begingroup$

Try $$f(x) = begin{cases} sin(1/x) &, x neq 0 \ 2 &, x = 0 end{cases} text{.} $$ What should $limsup_{x rightarrow 0} f(x)$ be?



Or to put this another way, why do you think $x_0$ cannot be a point in the sequence implicitly encoded by $x rightarrow x_0$? We have to go to some pains to dodge this point in the one-sided limits. (We don't take sequences of points, we take sequences of offsets, $h$ all required to be of the same sign, hence nonzero.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The lim sup on each side is 1, so it would make sense for the "general" lim sup to be the greater of the two, which is 1. My book claims the non-sided lim sup should also take $f(0)$ into account, so instead of 1 it's 2. I don't understand why this is preferred.
    $endgroup$
    – hiroshin
    Dec 5 '18 at 21:59










  • $begingroup$
    @hiroshin : How do you feel about the sequence of inputs to $f$: $(x_0, x_0, x_0, dots, x_0, dots)$? Is this a sequence we should use in our lim sup? What value does it put in the set whose supremum we are taking?
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 23:52
















1












1








1





$begingroup$

Try $$f(x) = begin{cases} sin(1/x) &, x neq 0 \ 2 &, x = 0 end{cases} text{.} $$ What should $limsup_{x rightarrow 0} f(x)$ be?



Or to put this another way, why do you think $x_0$ cannot be a point in the sequence implicitly encoded by $x rightarrow x_0$? We have to go to some pains to dodge this point in the one-sided limits. (We don't take sequences of points, we take sequences of offsets, $h$ all required to be of the same sign, hence nonzero.)






share|cite|improve this answer









$endgroup$



Try $$f(x) = begin{cases} sin(1/x) &, x neq 0 \ 2 &, x = 0 end{cases} text{.} $$ What should $limsup_{x rightarrow 0} f(x)$ be?



Or to put this another way, why do you think $x_0$ cannot be a point in the sequence implicitly encoded by $x rightarrow x_0$? We have to go to some pains to dodge this point in the one-sided limits. (We don't take sequences of points, we take sequences of offsets, $h$ all required to be of the same sign, hence nonzero.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 21:41









Eric TowersEric Towers

32.5k22369




32.5k22369












  • $begingroup$
    The lim sup on each side is 1, so it would make sense for the "general" lim sup to be the greater of the two, which is 1. My book claims the non-sided lim sup should also take $f(0)$ into account, so instead of 1 it's 2. I don't understand why this is preferred.
    $endgroup$
    – hiroshin
    Dec 5 '18 at 21:59










  • $begingroup$
    @hiroshin : How do you feel about the sequence of inputs to $f$: $(x_0, x_0, x_0, dots, x_0, dots)$? Is this a sequence we should use in our lim sup? What value does it put in the set whose supremum we are taking?
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 23:52




















  • $begingroup$
    The lim sup on each side is 1, so it would make sense for the "general" lim sup to be the greater of the two, which is 1. My book claims the non-sided lim sup should also take $f(0)$ into account, so instead of 1 it's 2. I don't understand why this is preferred.
    $endgroup$
    – hiroshin
    Dec 5 '18 at 21:59










  • $begingroup$
    @hiroshin : How do you feel about the sequence of inputs to $f$: $(x_0, x_0, x_0, dots, x_0, dots)$? Is this a sequence we should use in our lim sup? What value does it put in the set whose supremum we are taking?
    $endgroup$
    – Eric Towers
    Dec 5 '18 at 23:52


















$begingroup$
The lim sup on each side is 1, so it would make sense for the "general" lim sup to be the greater of the two, which is 1. My book claims the non-sided lim sup should also take $f(0)$ into account, so instead of 1 it's 2. I don't understand why this is preferred.
$endgroup$
– hiroshin
Dec 5 '18 at 21:59




$begingroup$
The lim sup on each side is 1, so it would make sense for the "general" lim sup to be the greater of the two, which is 1. My book claims the non-sided lim sup should also take $f(0)$ into account, so instead of 1 it's 2. I don't understand why this is preferred.
$endgroup$
– hiroshin
Dec 5 '18 at 21:59












$begingroup$
@hiroshin : How do you feel about the sequence of inputs to $f$: $(x_0, x_0, x_0, dots, x_0, dots)$? Is this a sequence we should use in our lim sup? What value does it put in the set whose supremum we are taking?
$endgroup$
– Eric Towers
Dec 5 '18 at 23:52






$begingroup$
@hiroshin : How do you feel about the sequence of inputs to $f$: $(x_0, x_0, x_0, dots, x_0, dots)$? Is this a sequence we should use in our lim sup? What value does it put in the set whose supremum we are taking?
$endgroup$
– Eric Towers
Dec 5 '18 at 23:52




















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