Why does the function $ f(x) = x, x in (0, 1) $ not have maximum or minimum values? Why do we not use limits?
$begingroup$
My textbook says that this function does not have a maximum value because for any $ x $ I choose to be a point of maximum or point of minimum, we can always choose some other $ x $ right next to it such that $ f(x) $ is smaller or greater (as we require). The same reasoning is given in this question here on Math SE.
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0. In other words, we know that: $$ lim_{x to 1^+} x = 1 quad text{ and } quad lim_{x to 0^-} x = 0 $$
In that case, aren't the extremum values technically just 1 and 0? Is there a specific reason why we use limits elsewhere in math but not in this particular case?
limits derivatives maxima-minima
$endgroup$
add a comment |
$begingroup$
My textbook says that this function does not have a maximum value because for any $ x $ I choose to be a point of maximum or point of minimum, we can always choose some other $ x $ right next to it such that $ f(x) $ is smaller or greater (as we require). The same reasoning is given in this question here on Math SE.
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0. In other words, we know that: $$ lim_{x to 1^+} x = 1 quad text{ and } quad lim_{x to 0^-} x = 0 $$
In that case, aren't the extremum values technically just 1 and 0? Is there a specific reason why we use limits elsewhere in math but not in this particular case?
limits derivatives maxima-minima
$endgroup$
1
$begingroup$
The limiting cases are used to define the infimum and supremum of the function.
$endgroup$
– Thomas Shelby
Jan 1 at 12:00
$begingroup$
Since (0,1) is a "open" interval, the reasoning sounds good. The limit of a function represents an approximate value not an exact value if the point is outside the domain of the function. That is, x=1 is not part of (0,1), so we can't say zero is a minimum.
$endgroup$
– NoChance
Jan 1 at 12:01
2
$begingroup$
We do take the limits, but instead of calling them the "maximum" and "minimum" we call them the "supremum" and "infimum". Tthe difference between (speaking loosely) a "maximum that is attained" and a "maximum that is only approached" is considered important enough to use different names for them.
$endgroup$
– bof
Jan 1 at 12:03
add a comment |
$begingroup$
My textbook says that this function does not have a maximum value because for any $ x $ I choose to be a point of maximum or point of minimum, we can always choose some other $ x $ right next to it such that $ f(x) $ is smaller or greater (as we require). The same reasoning is given in this question here on Math SE.
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0. In other words, we know that: $$ lim_{x to 1^+} x = 1 quad text{ and } quad lim_{x to 0^-} x = 0 $$
In that case, aren't the extremum values technically just 1 and 0? Is there a specific reason why we use limits elsewhere in math but not in this particular case?
limits derivatives maxima-minima
$endgroup$
My textbook says that this function does not have a maximum value because for any $ x $ I choose to be a point of maximum or point of minimum, we can always choose some other $ x $ right next to it such that $ f(x) $ is smaller or greater (as we require). The same reasoning is given in this question here on Math SE.
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0. In other words, we know that: $$ lim_{x to 1^+} x = 1 quad text{ and } quad lim_{x to 0^-} x = 0 $$
In that case, aren't the extremum values technically just 1 and 0? Is there a specific reason why we use limits elsewhere in math but not in this particular case?
limits derivatives maxima-minima
limits derivatives maxima-minima
asked Jan 1 at 11:55
WorldGovWorldGov
29810
29810
1
$begingroup$
The limiting cases are used to define the infimum and supremum of the function.
$endgroup$
– Thomas Shelby
Jan 1 at 12:00
$begingroup$
Since (0,1) is a "open" interval, the reasoning sounds good. The limit of a function represents an approximate value not an exact value if the point is outside the domain of the function. That is, x=1 is not part of (0,1), so we can't say zero is a minimum.
$endgroup$
– NoChance
Jan 1 at 12:01
2
$begingroup$
We do take the limits, but instead of calling them the "maximum" and "minimum" we call them the "supremum" and "infimum". Tthe difference between (speaking loosely) a "maximum that is attained" and a "maximum that is only approached" is considered important enough to use different names for them.
$endgroup$
– bof
Jan 1 at 12:03
add a comment |
1
$begingroup$
The limiting cases are used to define the infimum and supremum of the function.
$endgroup$
– Thomas Shelby
Jan 1 at 12:00
$begingroup$
Since (0,1) is a "open" interval, the reasoning sounds good. The limit of a function represents an approximate value not an exact value if the point is outside the domain of the function. That is, x=1 is not part of (0,1), so we can't say zero is a minimum.
$endgroup$
– NoChance
Jan 1 at 12:01
2
$begingroup$
We do take the limits, but instead of calling them the "maximum" and "minimum" we call them the "supremum" and "infimum". Tthe difference between (speaking loosely) a "maximum that is attained" and a "maximum that is only approached" is considered important enough to use different names for them.
$endgroup$
– bof
Jan 1 at 12:03
1
1
$begingroup$
The limiting cases are used to define the infimum and supremum of the function.
$endgroup$
– Thomas Shelby
Jan 1 at 12:00
$begingroup$
The limiting cases are used to define the infimum and supremum of the function.
$endgroup$
– Thomas Shelby
Jan 1 at 12:00
$begingroup$
Since (0,1) is a "open" interval, the reasoning sounds good. The limit of a function represents an approximate value not an exact value if the point is outside the domain of the function. That is, x=1 is not part of (0,1), so we can't say zero is a minimum.
$endgroup$
– NoChance
Jan 1 at 12:01
$begingroup$
Since (0,1) is a "open" interval, the reasoning sounds good. The limit of a function represents an approximate value not an exact value if the point is outside the domain of the function. That is, x=1 is not part of (0,1), so we can't say zero is a minimum.
$endgroup$
– NoChance
Jan 1 at 12:01
2
2
$begingroup$
We do take the limits, but instead of calling them the "maximum" and "minimum" we call them the "supremum" and "infimum". Tthe difference between (speaking loosely) a "maximum that is attained" and a "maximum that is only approached" is considered important enough to use different names for them.
$endgroup$
– bof
Jan 1 at 12:03
$begingroup$
We do take the limits, but instead of calling them the "maximum" and "minimum" we call them the "supremum" and "infimum". Tthe difference between (speaking loosely) a "maximum that is attained" and a "maximum that is only approached" is considered important enough to use different names for them.
$endgroup$
– bof
Jan 1 at 12:03
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The answer is that the definition of maximum and minimum of a set of real numbers $S$ (usually $x in S$ such that $forall y in S: x ge y$ for maximum) does not contain any limits, so they cannot be used to determined the maximum.
However, there is the definition of supremum and infimum (https://en.wikipedia.org/wiki/Infimum_and_supremum) that gets the answers 'you want'. Both definitions exist, because for some problems the maximum/minimum is important, while for other cases the supremum/infimum.
$endgroup$
add a comment |
$begingroup$
The issue about "having" minimum or maximum values should really be cast instead as "attaining" minimum or maximum values.
For your particular question, certainly $0$ is a lower found for $f(x)=x$ on the interval $(0,1)$, just as $1$ is an upper bound. But you cannot attain those values, as $0$ and $1$ are not in your domain! In fact, for any element $xin(0,1)$ there are always $y,zin(0,1)$ such that $y<x<z$.
The reason why we don't use limits to figure out whether or not we attain extreme values is simple -- limits ignore the point you are tending to! Recall your definition of a limit:
$$
forallepsilon>0,existsdelta>0text{ such that } (0<|x-c|<delta) Rightarrow (|f(x)-L|<epsilon)
$$
We are not allowed to let $x=c$ in the limit, and so the limiting value of a function need not be attained.
$endgroup$
add a comment |
$begingroup$
Because $0$ and $1$ are not in the domain of the function.
In general, consider a function $f$ which has domain $A$. A minimum or maximum of $f$ must be an element of $A$, i.e. $xin A$.
In your example $1notin (0,1)$ and $0notin (0,1)$.
On a side note, what you're referring to are denoted infimum and supremum.
$endgroup$
add a comment |
$begingroup$
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0.
The issue: $0$ and $1$ are not in the range of the function $f(x) = x$ when $x in (0,1)$. By definition, the maximum and minimum are in the range of the function: that's pretty much all there is to say about it.
What you touch on is the notion of infimum and supremum with respect to this function, which is slightly different. (Namely, the infimum is the largest $gamma$ such that $gamma leq f(x)$ all $x$ in the domain, and similarly supremum is the smallest $gamma$ such that $gamma geq f(x)$ for all $x$ in the domain.) This sort of limiting behavior you consider can be useful in finding the suprema/infima.
$endgroup$
$begingroup$
In your last parenthesis, it should be $cle f(x)$(or some other constant)(the same about the other inequality)
$endgroup$
– Holo
Jan 1 at 12:58
$begingroup$
Thanks for that, I derped a bit. (Common theme for me tonight. >_<)
$endgroup$
– Eevee Trainer
Jan 1 at 13:07
add a comment |
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4 Answers
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4 Answers
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$begingroup$
The answer is that the definition of maximum and minimum of a set of real numbers $S$ (usually $x in S$ such that $forall y in S: x ge y$ for maximum) does not contain any limits, so they cannot be used to determined the maximum.
However, there is the definition of supremum and infimum (https://en.wikipedia.org/wiki/Infimum_and_supremum) that gets the answers 'you want'. Both definitions exist, because for some problems the maximum/minimum is important, while for other cases the supremum/infimum.
$endgroup$
add a comment |
$begingroup$
The answer is that the definition of maximum and minimum of a set of real numbers $S$ (usually $x in S$ such that $forall y in S: x ge y$ for maximum) does not contain any limits, so they cannot be used to determined the maximum.
However, there is the definition of supremum and infimum (https://en.wikipedia.org/wiki/Infimum_and_supremum) that gets the answers 'you want'. Both definitions exist, because for some problems the maximum/minimum is important, while for other cases the supremum/infimum.
$endgroup$
add a comment |
$begingroup$
The answer is that the definition of maximum and minimum of a set of real numbers $S$ (usually $x in S$ such that $forall y in S: x ge y$ for maximum) does not contain any limits, so they cannot be used to determined the maximum.
However, there is the definition of supremum and infimum (https://en.wikipedia.org/wiki/Infimum_and_supremum) that gets the answers 'you want'. Both definitions exist, because for some problems the maximum/minimum is important, while for other cases the supremum/infimum.
$endgroup$
The answer is that the definition of maximum and minimum of a set of real numbers $S$ (usually $x in S$ such that $forall y in S: x ge y$ for maximum) does not contain any limits, so they cannot be used to determined the maximum.
However, there is the definition of supremum and infimum (https://en.wikipedia.org/wiki/Infimum_and_supremum) that gets the answers 'you want'. Both definitions exist, because for some problems the maximum/minimum is important, while for other cases the supremum/infimum.
answered Jan 1 at 12:03
IngixIngix
3,464146
3,464146
add a comment |
add a comment |
$begingroup$
The issue about "having" minimum or maximum values should really be cast instead as "attaining" minimum or maximum values.
For your particular question, certainly $0$ is a lower found for $f(x)=x$ on the interval $(0,1)$, just as $1$ is an upper bound. But you cannot attain those values, as $0$ and $1$ are not in your domain! In fact, for any element $xin(0,1)$ there are always $y,zin(0,1)$ such that $y<x<z$.
The reason why we don't use limits to figure out whether or not we attain extreme values is simple -- limits ignore the point you are tending to! Recall your definition of a limit:
$$
forallepsilon>0,existsdelta>0text{ such that } (0<|x-c|<delta) Rightarrow (|f(x)-L|<epsilon)
$$
We are not allowed to let $x=c$ in the limit, and so the limiting value of a function need not be attained.
$endgroup$
add a comment |
$begingroup$
The issue about "having" minimum or maximum values should really be cast instead as "attaining" minimum or maximum values.
For your particular question, certainly $0$ is a lower found for $f(x)=x$ on the interval $(0,1)$, just as $1$ is an upper bound. But you cannot attain those values, as $0$ and $1$ are not in your domain! In fact, for any element $xin(0,1)$ there are always $y,zin(0,1)$ such that $y<x<z$.
The reason why we don't use limits to figure out whether or not we attain extreme values is simple -- limits ignore the point you are tending to! Recall your definition of a limit:
$$
forallepsilon>0,existsdelta>0text{ such that } (0<|x-c|<delta) Rightarrow (|f(x)-L|<epsilon)
$$
We are not allowed to let $x=c$ in the limit, and so the limiting value of a function need not be attained.
$endgroup$
add a comment |
$begingroup$
The issue about "having" minimum or maximum values should really be cast instead as "attaining" minimum or maximum values.
For your particular question, certainly $0$ is a lower found for $f(x)=x$ on the interval $(0,1)$, just as $1$ is an upper bound. But you cannot attain those values, as $0$ and $1$ are not in your domain! In fact, for any element $xin(0,1)$ there are always $y,zin(0,1)$ such that $y<x<z$.
The reason why we don't use limits to figure out whether or not we attain extreme values is simple -- limits ignore the point you are tending to! Recall your definition of a limit:
$$
forallepsilon>0,existsdelta>0text{ such that } (0<|x-c|<delta) Rightarrow (|f(x)-L|<epsilon)
$$
We are not allowed to let $x=c$ in the limit, and so the limiting value of a function need not be attained.
$endgroup$
The issue about "having" minimum or maximum values should really be cast instead as "attaining" minimum or maximum values.
For your particular question, certainly $0$ is a lower found for $f(x)=x$ on the interval $(0,1)$, just as $1$ is an upper bound. But you cannot attain those values, as $0$ and $1$ are not in your domain! In fact, for any element $xin(0,1)$ there are always $y,zin(0,1)$ such that $y<x<z$.
The reason why we don't use limits to figure out whether or not we attain extreme values is simple -- limits ignore the point you are tending to! Recall your definition of a limit:
$$
forallepsilon>0,existsdelta>0text{ such that } (0<|x-c|<delta) Rightarrow (|f(x)-L|<epsilon)
$$
We are not allowed to let $x=c$ in the limit, and so the limiting value of a function need not be attained.
answered Jan 1 at 12:01
InequalitiesEverywhereInequalitiesEverywhere
1313
1313
add a comment |
add a comment |
$begingroup$
Because $0$ and $1$ are not in the domain of the function.
In general, consider a function $f$ which has domain $A$. A minimum or maximum of $f$ must be an element of $A$, i.e. $xin A$.
In your example $1notin (0,1)$ and $0notin (0,1)$.
On a side note, what you're referring to are denoted infimum and supremum.
$endgroup$
add a comment |
$begingroup$
Because $0$ and $1$ are not in the domain of the function.
In general, consider a function $f$ which has domain $A$. A minimum or maximum of $f$ must be an element of $A$, i.e. $xin A$.
In your example $1notin (0,1)$ and $0notin (0,1)$.
On a side note, what you're referring to are denoted infimum and supremum.
$endgroup$
add a comment |
$begingroup$
Because $0$ and $1$ are not in the domain of the function.
In general, consider a function $f$ which has domain $A$. A minimum or maximum of $f$ must be an element of $A$, i.e. $xin A$.
In your example $1notin (0,1)$ and $0notin (0,1)$.
On a side note, what you're referring to are denoted infimum and supremum.
$endgroup$
Because $0$ and $1$ are not in the domain of the function.
In general, consider a function $f$ which has domain $A$. A minimum or maximum of $f$ must be an element of $A$, i.e. $xin A$.
In your example $1notin (0,1)$ and $0notin (0,1)$.
On a side note, what you're referring to are denoted infimum and supremum.
answered Jan 1 at 12:01
EffEff
11.5k21638
11.5k21638
add a comment |
add a comment |
$begingroup$
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0.
The issue: $0$ and $1$ are not in the range of the function $f(x) = x$ when $x in (0,1)$. By definition, the maximum and minimum are in the range of the function: that's pretty much all there is to say about it.
What you touch on is the notion of infimum and supremum with respect to this function, which is slightly different. (Namely, the infimum is the largest $gamma$ such that $gamma leq f(x)$ all $x$ in the domain, and similarly supremum is the smallest $gamma$ such that $gamma geq f(x)$ for all $x$ in the domain.) This sort of limiting behavior you consider can be useful in finding the suprema/infima.
$endgroup$
$begingroup$
In your last parenthesis, it should be $cle f(x)$(or some other constant)(the same about the other inequality)
$endgroup$
– Holo
Jan 1 at 12:58
$begingroup$
Thanks for that, I derped a bit. (Common theme for me tonight. >_<)
$endgroup$
– Eevee Trainer
Jan 1 at 13:07
add a comment |
$begingroup$
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0.
The issue: $0$ and $1$ are not in the range of the function $f(x) = x$ when $x in (0,1)$. By definition, the maximum and minimum are in the range of the function: that's pretty much all there is to say about it.
What you touch on is the notion of infimum and supremum with respect to this function, which is slightly different. (Namely, the infimum is the largest $gamma$ such that $gamma leq f(x)$ all $x$ in the domain, and similarly supremum is the smallest $gamma$ such that $gamma geq f(x)$ for all $x$ in the domain.) This sort of limiting behavior you consider can be useful in finding the suprema/infima.
$endgroup$
$begingroup$
In your last parenthesis, it should be $cle f(x)$(or some other constant)(the same about the other inequality)
$endgroup$
– Holo
Jan 1 at 12:58
$begingroup$
Thanks for that, I derped a bit. (Common theme for me tonight. >_<)
$endgroup$
– Eevee Trainer
Jan 1 at 13:07
add a comment |
$begingroup$
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0.
The issue: $0$ and $1$ are not in the range of the function $f(x) = x$ when $x in (0,1)$. By definition, the maximum and minimum are in the range of the function: that's pretty much all there is to say about it.
What you touch on is the notion of infimum and supremum with respect to this function, which is slightly different. (Namely, the infimum is the largest $gamma$ such that $gamma leq f(x)$ all $x$ in the domain, and similarly supremum is the smallest $gamma$ such that $gamma geq f(x)$ for all $x$ in the domain.) This sort of limiting behavior you consider can be useful in finding the suprema/infima.
$endgroup$
My question is, why don't we use limits here and say that the maximum value is just 1, and that the minimum value is just 0.
The issue: $0$ and $1$ are not in the range of the function $f(x) = x$ when $x in (0,1)$. By definition, the maximum and minimum are in the range of the function: that's pretty much all there is to say about it.
What you touch on is the notion of infimum and supremum with respect to this function, which is slightly different. (Namely, the infimum is the largest $gamma$ such that $gamma leq f(x)$ all $x$ in the domain, and similarly supremum is the smallest $gamma$ such that $gamma geq f(x)$ for all $x$ in the domain.) This sort of limiting behavior you consider can be useful in finding the suprema/infima.
edited Jan 1 at 13:07
answered Jan 1 at 12:04
Eevee TrainerEevee Trainer
5,7471936
5,7471936
$begingroup$
In your last parenthesis, it should be $cle f(x)$(or some other constant)(the same about the other inequality)
$endgroup$
– Holo
Jan 1 at 12:58
$begingroup$
Thanks for that, I derped a bit. (Common theme for me tonight. >_<)
$endgroup$
– Eevee Trainer
Jan 1 at 13:07
add a comment |
$begingroup$
In your last parenthesis, it should be $cle f(x)$(or some other constant)(the same about the other inequality)
$endgroup$
– Holo
Jan 1 at 12:58
$begingroup$
Thanks for that, I derped a bit. (Common theme for me tonight. >_<)
$endgroup$
– Eevee Trainer
Jan 1 at 13:07
$begingroup$
In your last parenthesis, it should be $cle f(x)$(or some other constant)(the same about the other inequality)
$endgroup$
– Holo
Jan 1 at 12:58
$begingroup$
In your last parenthesis, it should be $cle f(x)$(or some other constant)(the same about the other inequality)
$endgroup$
– Holo
Jan 1 at 12:58
$begingroup$
Thanks for that, I derped a bit. (Common theme for me tonight. >_<)
$endgroup$
– Eevee Trainer
Jan 1 at 13:07
$begingroup$
Thanks for that, I derped a bit. (Common theme for me tonight. >_<)
$endgroup$
– Eevee Trainer
Jan 1 at 13:07
add a comment |
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$begingroup$
The limiting cases are used to define the infimum and supremum of the function.
$endgroup$
– Thomas Shelby
Jan 1 at 12:00
$begingroup$
Since (0,1) is a "open" interval, the reasoning sounds good. The limit of a function represents an approximate value not an exact value if the point is outside the domain of the function. That is, x=1 is not part of (0,1), so we can't say zero is a minimum.
$endgroup$
– NoChance
Jan 1 at 12:01
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We do take the limits, but instead of calling them the "maximum" and "minimum" we call them the "supremum" and "infimum". Tthe difference between (speaking loosely) a "maximum that is attained" and a "maximum that is only approached" is considered important enough to use different names for them.
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– bof
Jan 1 at 12:03