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Making sense of limits at infinity.
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I want to define things like $lim_x to +infty f(x)$ and $lim_x to a f(x) = + infty$ for a function $f: E subseteq mathbbR to mathbbR$.
In particular, when is it necessary that the point at which the limit point is taken is a limit point (it is necessary if this point is real, but what about the infinite case?).
What conditions must be imposed on the domain $E$? Does it need to be bounded in the infinite case.
Many thanks for any insights. I'm looking for a definition that is generally applicable.
calculus real-analysis
asked Jul 29 at 7:56
Math_QED
6,34631344
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up vote
0
down vote
favorite
I want to define things like $lim_x to +infty f(x)$ and $lim_x to a f(x) = + infty$ for a function $f: E subseteq mathbbR to mathbbR$.
In particular, when is it necessary that the point at which the limit point is taken is a limit point (it is necessary if this point is real, but what about the infinite case?).
What conditions must be imposed on the domain $E$? Does it need to be bounded in the infinite case.
Many thanks for any insights. I'm looking for a definition that is generally applicable.
calculus real-analysis
asked Jul 29 at 7:56
Math_QED
6,34631344
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to define things like $lim_x to +infty f(x)$ and $lim_x to a f(x) = + infty$ for a function $f: E subseteq mathbbR to mathbbR$.
In particular, when is it necessary that the point at which the limit point is taken is a limit point (it is necessary if this point is real, but what about the infinite case?).
What conditions must be imposed on the domain $E$? Does it need to be bounded in the infinite case.
Many thanks for any insights. I'm looking for a definition that is generally applicable.
calculus real-analysis
asked Jul 29 at 7:56
Math_QED
6,34631344
I want to define things like $lim_x to +infty f(x)$ and $lim_x to a f(x) = + infty$ for a function $f: E subseteq mathbbR to mathbbR$.
In particular, when is it necessary that the point at which the limit point is taken is a limit point (it is necessary if this point is real, but what about the infinite case?).
What conditions must be imposed on the domain $E$? Does it need to be bounded in the infinite case.
Many thanks for any insights. I'm looking for a definition that is generally applicable.
calculus real-analysis
asked Jul 29 at 7:56
Math_QED
6,34631344
asked Jul 29 at 7:56
Math_QED
6,34631344
asked Jul 29 at 7:56
Math_QED
6,34631344
asked Jul 29 at 7:56
Math_QED
6,34631344
6,34631344
add a comment |Â
add a comment |Â
3 Answers
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active
oldest
votes
up vote
1
down vote
Limits involving $pm infty$ are best understood as being ordinary limits of functions domain or codomain is a subset of extended real numbers, in which case the values $+infty$ and $-infty$ simply refer to ordinary points of said topological space and can be treated as any other.
In a typical basis for the topology on the extended real numbers, the neighborhoods of $+infty$ are the intervals $(a, +infty]$, and similarly for $-infty$.
A subset of the reals, viewed as a subset of the extended reals, has $+infty$ as a limit point if and only if it is unbounded in the positive direction.
The extended real numbers are even metrizable if you really want to use metric space methods rather than general topological ones. Of course, to do so you can't pick a metric extending the usual $d(a,b) = |a-b|$ metric.
answered Jul 29 at 10:22
Hurkyl
107k9112253
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up vote
1
down vote
One quite flexible generalization of limits is to work with filters.
If $f:Eto C$ where $E$ and $C$ are arbitrary sets, and $mathcal F$ and $mathcal G$ are filters on $E$ and $C$, respectively, we can define
Say that $f(x)to mathcal G$ as $xto mathcal F$ (here $to$ is pronounced "goes to") if one of the following equivalent conditions hold:
- For every $Ginmathcal G$ there is an $Finmathcal F$ such that $f(F)subseteq G$.
- For every $Ginmathcal G$, it holds that $f^-1(G)inmathcal F$.
If $E$ and $C$ are topological spaces and we choose $mathcal F$ as a punctured neighborhood filter and $mathcal G$ as a neighborhood filter, then this produces the usual notion of limit.
Different choices of $mathcal F$ can encode different limits -- such as a left neighborhood filter for "$x$ approaches $x_0$ from above", or the filter of all supersets of $[n,infty)$ for some $n$ to give "$xto+infty$".
Similarly, choices of $mathcal G$ can vary what we claims happens to $f(x)$, such as -- again -- the filter of all supersets of $[n,infty)$ to say $f(x)to+infty$, or the filter of all complements of bounded subsets of $mathbb R$ for $f(x)topminfty$. Or even things we don't usually see in calculus, e.g. a one-sided neighborhood filter for "approaches $y_0$ from below", or a punctured neighborhood filter for "approaches but does not reach".
If $E$ is a proper subset of $mathbb R$ (that is, $f(x)$ is not defined for all real $x$), then we can construct an $mathcal F$ by taking a filter on $mathbb R$ and intersecting all of its elements with $E$. This expresses what happens for $lim_xto af(x)$ when $a$ is not in the domain of $f$: Take the neighborhood filter at $a$ in $mathbb R$, and restrict it to the domain of $f$.
The condition that $a$ must be a limit point corresponds to demanding that $varnothingnotinmathcal F$, which is built into the definition of filter. Or in other words, the "intersect everything with $E$" construction will not produce a filter if the complement of $E$ is in the filter on $mathbb R$ we start with.
The same construction can produce the filters for $xto+infty$ and $xto-infty$ by intersecting with neighborhood filters on the extended real line (as Hurkyl explains).
This also produces $ntoinfty$ limits for sequences -- that is, functions $mathbb Ntomathbb R$.
answered Jul 29 at 11:33
Henning Makholm
225k16290516
add a comment |Â
up vote
0
down vote
For $lim_x to +infty f(x)$, the suitable requirement is that $E$ is not bounded above, i.e., for every $omega in mathbbR$ there exists $x in E$ such that $x > omega$.
(Since if $E$ is bounded above, then $f(x)$ is undefined for all sufficiently large $x$, and then it's meaningless to ask if $f(x)$ tends to something as $x to +infty$.)
answered Jul 29 at 8:03
Hans Lundmark
32.7k563109
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Limits involving $pm infty$ are best understood as being ordinary limits of functions domain or codomain is a subset of extended real numbers, in which case the values $+infty$ and $-infty$ simply refer to ordinary points of said topological space and can be treated as any other.
In a typical basis for the topology on the extended real numbers, the neighborhoods of $+infty$ are the intervals $(a, +infty]$, and similarly for $-infty$.
A subset of the reals, viewed as a subset of the extended reals, has $+infty$ as a limit point if and only if it is unbounded in the positive direction.
The extended real numbers are even metrizable if you really want to use metric space methods rather than general topological ones. Of course, to do so you can't pick a metric extending the usual $d(a,b) = |a-b|$ metric.
answered Jul 29 at 10:22
Hurkyl
107k9112253
add a comment |Â
up vote
1
down vote
Limits involving $pm infty$ are best understood as being ordinary limits of functions domain or codomain is a subset of extended real numbers, in which case the values $+infty$ and $-infty$ simply refer to ordinary points of said topological space and can be treated as any other.
In a typical basis for the topology on the extended real numbers, the neighborhoods of $+infty$ are the intervals $(a, +infty]$, and similarly for $-infty$.
A subset of the reals, viewed as a subset of the extended reals, has $+infty$ as a limit point if and only if it is unbounded in the positive direction.
The extended real numbers are even metrizable if you really want to use metric space methods rather than general topological ones. Of course, to do so you can't pick a metric extending the usual $d(a,b) = |a-b|$ metric.
answered Jul 29 at 10:22
Hurkyl
107k9112253
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Limits involving $pm infty$ are best understood as being ordinary limits of functions domain or codomain is a subset of extended real numbers, in which case the values $+infty$ and $-infty$ simply refer to ordinary points of said topological space and can be treated as any other.
In a typical basis for the topology on the extended real numbers, the neighborhoods of $+infty$ are the intervals $(a, +infty]$, and similarly for $-infty$.
A subset of the reals, viewed as a subset of the extended reals, has $+infty$ as a limit point if and only if it is unbounded in the positive direction.
The extended real numbers are even metrizable if you really want to use metric space methods rather than general topological ones. Of course, to do so you can't pick a metric extending the usual $d(a,b) = |a-b|$ metric.
answered Jul 29 at 10:22
Hurkyl
107k9112253
Limits involving $pm infty$ are best understood as being ordinary limits of functions domain or codomain is a subset of extended real numbers, in which case the values $+infty$ and $-infty$ simply refer to ordinary points of said topological space and can be treated as any other.
In a typical basis for the topology on the extended real numbers, the neighborhoods of $+infty$ are the intervals $(a, +infty]$, and similarly for $-infty$.
A subset of the reals, viewed as a subset of the extended reals, has $+infty$ as a limit point if and only if it is unbounded in the positive direction.
The extended real numbers are even metrizable if you really want to use metric space methods rather than general topological ones. Of course, to do so you can't pick a metric extending the usual $d(a,b) = |a-b|$ metric.
answered Jul 29 at 10:22
Hurkyl
107k9112253
edited Jul 29 at 10:28
answered Jul 29 at 10:22
Hurkyl
107k9112253
answered Jul 29 at 10:22
Hurkyl
107k9112253
answered Jul 29 at 10:22
Hurkyl
107k9112253
107k9112253
add a comment |Â
add a comment |Â
up vote
1
down vote
One quite flexible generalization of limits is to work with filters.
If $f:Eto C$ where $E$ and $C$ are arbitrary sets, and $mathcal F$ and $mathcal G$ are filters on $E$ and $C$, respectively, we can define
Say that $f(x)to mathcal G$ as $xto mathcal F$ (here $to$ is pronounced "goes to") if one of the following equivalent conditions hold:
- For every $Ginmathcal G$ there is an $Finmathcal F$ such that $f(F)subseteq G$.
- For every $Ginmathcal G$, it holds that $f^-1(G)inmathcal F$.
If $E$ and $C$ are topological spaces and we choose $mathcal F$ as a punctured neighborhood filter and $mathcal G$ as a neighborhood filter, then this produces the usual notion of limit.
Different choices of $mathcal F$ can encode different limits -- such as a left neighborhood filter for "$x$ approaches $x_0$ from above", or the filter of all supersets of $[n,infty)$ for some $n$ to give "$xto+infty$".
Similarly, choices of $mathcal G$ can vary what we claims happens to $f(x)$, such as -- again -- the filter of all supersets of $[n,infty)$ to say $f(x)to+infty$, or the filter of all complements of bounded subsets of $mathbb R$ for $f(x)topminfty$. Or even things we don't usually see in calculus, e.g. a one-sided neighborhood filter for "approaches $y_0$ from below", or a punctured neighborhood filter for "approaches but does not reach".
If $E$ is a proper subset of $mathbb R$ (that is, $f(x)$ is not defined for all real $x$), then we can construct an $mathcal F$ by taking a filter on $mathbb R$ and intersecting all of its elements with $E$. This expresses what happens for $lim_xto af(x)$ when $a$ is not in the domain of $f$: Take the neighborhood filter at $a$ in $mathbb R$, and restrict it to the domain of $f$.
The condition that $a$ must be a limit point corresponds to demanding that $varnothingnotinmathcal F$, which is built into the definition of filter. Or in other words, the "intersect everything with $E$" construction will not produce a filter if the complement of $E$ is in the filter on $mathbb R$ we start with.
The same construction can produce the filters for $xto+infty$ and $xto-infty$ by intersecting with neighborhood filters on the extended real line (as Hurkyl explains).
This also produces $ntoinfty$ limits for sequences -- that is, functions $mathbb Ntomathbb R$.
answered Jul 29 at 11:33
Henning Makholm
225k16290516
add a comment |Â
up vote
1
down vote
One quite flexible generalization of limits is to work with filters.
If $f:Eto C$ where $E$ and $C$ are arbitrary sets, and $mathcal F$ and $mathcal G$ are filters on $E$ and $C$, respectively, we can define
Say that $f(x)to mathcal G$ as $xto mathcal F$ (here $to$ is pronounced "goes to") if one of the following equivalent conditions hold:
- For every $Ginmathcal G$ there is an $Finmathcal F$ such that $f(F)subseteq G$.
- For every $Ginmathcal G$, it holds that $f^-1(G)inmathcal F$.
If $E$ and $C$ are topological spaces and we choose $mathcal F$ as a punctured neighborhood filter and $mathcal G$ as a neighborhood filter, then this produces the usual notion of limit.
Different choices of $mathcal F$ can encode different limits -- such as a left neighborhood filter for "$x$ approaches $x_0$ from above", or the filter of all supersets of $[n,infty)$ for some $n$ to give "$xto+infty$".
Similarly, choices of $mathcal G$ can vary what we claims happens to $f(x)$, such as -- again -- the filter of all supersets of $[n,infty)$ to say $f(x)to+infty$, or the filter of all complements of bounded subsets of $mathbb R$ for $f(x)topminfty$. Or even things we don't usually see in calculus, e.g. a one-sided neighborhood filter for "approaches $y_0$ from below", or a punctured neighborhood filter for "approaches but does not reach".
If $E$ is a proper subset of $mathbb R$ (that is, $f(x)$ is not defined for all real $x$), then we can construct an $mathcal F$ by taking a filter on $mathbb R$ and intersecting all of its elements with $E$. This expresses what happens for $lim_xto af(x)$ when $a$ is not in the domain of $f$: Take the neighborhood filter at $a$ in $mathbb R$, and restrict it to the domain of $f$.
The condition that $a$ must be a limit point corresponds to demanding that $varnothingnotinmathcal F$, which is built into the definition of filter. Or in other words, the "intersect everything with $E$" construction will not produce a filter if the complement of $E$ is in the filter on $mathbb R$ we start with.
The same construction can produce the filters for $xto+infty$ and $xto-infty$ by intersecting with neighborhood filters on the extended real line (as Hurkyl explains).
This also produces $ntoinfty$ limits for sequences -- that is, functions $mathbb Ntomathbb R$.
answered Jul 29 at 11:33
Henning Makholm
225k16290516
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One quite flexible generalization of limits is to work with filters.
If $f:Eto C$ where $E$ and $C$ are arbitrary sets, and $mathcal F$ and $mathcal G$ are filters on $E$ and $C$, respectively, we can define
Say that $f(x)to mathcal G$ as $xto mathcal F$ (here $to$ is pronounced "goes to") if one of the following equivalent conditions hold:
- For every $Ginmathcal G$ there is an $Finmathcal F$ such that $f(F)subseteq G$.
- For every $Ginmathcal G$, it holds that $f^-1(G)inmathcal F$.
If $E$ and $C$ are topological spaces and we choose $mathcal F$ as a punctured neighborhood filter and $mathcal G$ as a neighborhood filter, then this produces the usual notion of limit.
Different choices of $mathcal F$ can encode different limits -- such as a left neighborhood filter for "$x$ approaches $x_0$ from above", or the filter of all supersets of $[n,infty)$ for some $n$ to give "$xto+infty$".
Similarly, choices of $mathcal G$ can vary what we claims happens to $f(x)$, such as -- again -- the filter of all supersets of $[n,infty)$ to say $f(x)to+infty$, or the filter of all complements of bounded subsets of $mathbb R$ for $f(x)topminfty$. Or even things we don't usually see in calculus, e.g. a one-sided neighborhood filter for "approaches $y_0$ from below", or a punctured neighborhood filter for "approaches but does not reach".
If $E$ is a proper subset of $mathbb R$ (that is, $f(x)$ is not defined for all real $x$), then we can construct an $mathcal F$ by taking a filter on $mathbb R$ and intersecting all of its elements with $E$. This expresses what happens for $lim_xto af(x)$ when $a$ is not in the domain of $f$: Take the neighborhood filter at $a$ in $mathbb R$, and restrict it to the domain of $f$.
The condition that $a$ must be a limit point corresponds to demanding that $varnothingnotinmathcal F$, which is built into the definition of filter. Or in other words, the "intersect everything with $E$" construction will not produce a filter if the complement of $E$ is in the filter on $mathbb R$ we start with.
The same construction can produce the filters for $xto+infty$ and $xto-infty$ by intersecting with neighborhood filters on the extended real line (as Hurkyl explains).
This also produces $ntoinfty$ limits for sequences -- that is, functions $mathbb Ntomathbb R$.
answered Jul 29 at 11:33
Henning Makholm
225k16290516
One quite flexible generalization of limits is to work with filters.
If $f:Eto C$ where $E$ and $C$ are arbitrary sets, and $mathcal F$ and $mathcal G$ are filters on $E$ and $C$, respectively, we can define
Say that $f(x)to mathcal G$ as $xto mathcal F$ (here $to$ is pronounced "goes to") if one of the following equivalent conditions hold:
- For every $Ginmathcal G$ there is an $Finmathcal F$ such that $f(F)subseteq G$.
- For every $Ginmathcal G$, it holds that $f^-1(G)inmathcal F$.
If $E$ and $C$ are topological spaces and we choose $mathcal F$ as a punctured neighborhood filter and $mathcal G$ as a neighborhood filter, then this produces the usual notion of limit.
Different choices of $mathcal F$ can encode different limits -- such as a left neighborhood filter for "$x$ approaches $x_0$ from above", or the filter of all supersets of $[n,infty)$ for some $n$ to give "$xto+infty$".
Similarly, choices of $mathcal G$ can vary what we claims happens to $f(x)$, such as -- again -- the filter of all supersets of $[n,infty)$ to say $f(x)to+infty$, or the filter of all complements of bounded subsets of $mathbb R$ for $f(x)topminfty$. Or even things we don't usually see in calculus, e.g. a one-sided neighborhood filter for "approaches $y_0$ from below", or a punctured neighborhood filter for "approaches but does not reach".
If $E$ is a proper subset of $mathbb R$ (that is, $f(x)$ is not defined for all real $x$), then we can construct an $mathcal F$ by taking a filter on $mathbb R$ and intersecting all of its elements with $E$. This expresses what happens for $lim_xto af(x)$ when $a$ is not in the domain of $f$: Take the neighborhood filter at $a$ in $mathbb R$, and restrict it to the domain of $f$.
The condition that $a$ must be a limit point corresponds to demanding that $varnothingnotinmathcal F$, which is built into the definition of filter. Or in other words, the "intersect everything with $E$" construction will not produce a filter if the complement of $E$ is in the filter on $mathbb R$ we start with.
The same construction can produce the filters for $xto+infty$ and $xto-infty$ by intersecting with neighborhood filters on the extended real line (as Hurkyl explains).
This also produces $ntoinfty$ limits for sequences -- that is, functions $mathbb Ntomathbb R$.
answered Jul 29 at 11:33
Henning Makholm
225k16290516
answered Jul 29 at 11:33
Henning Makholm
225k16290516
answered Jul 29 at 11:33
Henning Makholm
225k16290516
answered Jul 29 at 11:33
Henning Makholm
225k16290516
225k16290516
add a comment |Â
add a comment |Â
up vote
0
down vote
For $lim_x to +infty f(x)$, the suitable requirement is that $E$ is not bounded above, i.e., for every $omega in mathbbR$ there exists $x in E$ such that $x > omega$.
(Since if $E$ is bounded above, then $f(x)$ is undefined for all sufficiently large $x$, and then it's meaningless to ask if $f(x)$ tends to something as $x to +infty$.)
answered Jul 29 at 8:03
Hans Lundmark
32.7k563109
add a comment |Â
up vote
0
down vote
For $lim_x to +infty f(x)$, the suitable requirement is that $E$ is not bounded above, i.e., for every $omega in mathbbR$ there exists $x in E$ such that $x > omega$.
(Since if $E$ is bounded above, then $f(x)$ is undefined for all sufficiently large $x$, and then it's meaningless to ask if $f(x)$ tends to something as $x to +infty$.)
answered Jul 29 at 8:03
Hans Lundmark
32.7k563109
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For $lim_x to +infty f(x)$, the suitable requirement is that $E$ is not bounded above, i.e., for every $omega in mathbbR$ there exists $x in E$ such that $x > omega$.
(Since if $E$ is bounded above, then $f(x)$ is undefined for all sufficiently large $x$, and then it's meaningless to ask if $f(x)$ tends to something as $x to +infty$.)
answered Jul 29 at 8:03
Hans Lundmark
32.7k563109
For $lim_x to +infty f(x)$, the suitable requirement is that $E$ is not bounded above, i.e., for every $omega in mathbbR$ there exists $x in E$ such that $x > omega$.
(Since if $E$ is bounded above, then $f(x)$ is undefined for all sufficiently large $x$, and then it's meaningless to ask if $f(x)$ tends to something as $x to +infty$.)
answered Jul 29 at 8:03
Hans Lundmark
32.7k563109
answered Jul 29 at 8:03
Hans Lundmark
32.7k563109
answered Jul 29 at 8:03
Hans Lundmark
32.7k563109
answered Jul 29 at 8:03
Hans Lundmark
32.7k563109
32.7k563109
add a comment |Â
add a comment |Â
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