Mixing
How can we define the limit of a constant function?
$begingroup$
Wikipedia says:
In mathematics, a limit is the value that a function(or sequence) "approaches" as the input (or index) "approaches" some value.
What if the function was a constant?! A constant function will not approach anything, so, how would we define the limit of a constant function?
calculus limits
asked Sep 19 '18 at 2:17
MuhammadMuhammad
2118
$endgroup$
|
show 3 more comments
$begingroup$
Wikipedia says:
In mathematics, a limit is the value that a function(or sequence) "approaches" as the input (or index) "approaches" some value.
What if the function was a constant?! A constant function will not approach anything, so, how would we define the limit of a constant function?
calculus limits
asked Sep 19 '18 at 2:17
MuhammadMuhammad
2118
$endgroup$
1
$begingroup$
Limit of WHAT? There is no "input" to a constant. It is nonsense to write $lim_{x to x_0} pi$, for instance.
$endgroup$
– David G. Stork
Sep 19 '18 at 2:30
15
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David I would interpret your limit to have value $pi$.
$endgroup$
– M_B
Sep 19 '18 at 2:35
15
$begingroup$
@DavidG.Stork It's not nonsense to write $lim_{x to x_0} pi$, but it is trivial. Here, $pi = f(x)$ is a constant function, so the definition is perfectly valid here.
$endgroup$
– JavaMan
Sep 19 '18 at 2:43
$begingroup$
The value of a function $f(x) $ may equal its limit $L$ as $xto a$. Check for example $f(x) =xsin (1/x)$ as $xto 0$. For a constant function the function value always equals its limits.
$endgroup$
– Paramanand Singh
Sep 19 '18 at 3:11
$begingroup$
I have seen this question in regard to the value of periodic number 0.(9) . In this case, indeed, some people try to say that it's not equal to 1 but the "limit" of this number is 1, disregarding the fact that the number is constant.
$endgroup$
– IMil
Sep 19 '18 at 5:19
|
show 3 more comments
$begingroup$
Wikipedia says:
In mathematics, a limit is the value that a function(or sequence) "approaches" as the input (or index) "approaches" some value.
What if the function was a constant?! A constant function will not approach anything, so, how would we define the limit of a constant function?
calculus limits
asked Sep 19 '18 at 2:17
MuhammadMuhammad
2118
$endgroup$
Wikipedia says:
In mathematics, a limit is the value that a function(or sequence) "approaches" as the input (or index) "approaches" some value.
What if the function was a constant?! A constant function will not approach anything, so, how would we define the limit of a constant function?
calculus limits
calculus limits
asked Sep 19 '18 at 2:17
MuhammadMuhammad
2118
asked Sep 19 '18 at 2:17
MuhammadMuhammad
2118
edited Jan 11 at 7:54
Muhammad
asked Sep 19 '18 at 2:17
MuhammadMuhammad
2118
asked Sep 19 '18 at 2:17
MuhammadMuhammad
2118
asked Sep 19 '18 at 2:17
MuhammadMuhammad
2118
2118
1
$begingroup$
Limit of WHAT? There is no "input" to a constant. It is nonsense to write $lim_{x to x_0} pi$, for instance.
$endgroup$
– David G. Stork
Sep 19 '18 at 2:30
15
$begingroup$
David I would interpret your limit to have value $pi$.
$endgroup$
– M_B
Sep 19 '18 at 2:35
15
$begingroup$
@DavidG.Stork It's not nonsense to write $lim_{x to x_0} pi$, but it is trivial. Here, $pi = f(x)$ is a constant function, so the definition is perfectly valid here.
$endgroup$
– JavaMan
Sep 19 '18 at 2:43
$begingroup$
The value of a function $f(x) $ may equal its limit $L$ as $xto a$. Check for example $f(x) =xsin (1/x)$ as $xto 0$. For a constant function the function value always equals its limits.
$endgroup$
– Paramanand Singh
Sep 19 '18 at 3:11
$begingroup$
I have seen this question in regard to the value of periodic number 0.(9) . In this case, indeed, some people try to say that it's not equal to 1 but the "limit" of this number is 1, disregarding the fact that the number is constant.
$endgroup$
– IMil
Sep 19 '18 at 5:19
|
show 3 more comments
1
$begingroup$
Limit of WHAT? There is no "input" to a constant. It is nonsense to write $lim_{x to x_0} pi$, for instance.
$endgroup$
– David G. Stork
Sep 19 '18 at 2:30
15
$begingroup$
David I would interpret your limit to have value $pi$.
$endgroup$
– M_B
Sep 19 '18 at 2:35
15
$begingroup$
@DavidG.Stork It's not nonsense to write $lim_{x to x_0} pi$, but it is trivial. Here, $pi = f(x)$ is a constant function, so the definition is perfectly valid here.
$endgroup$
– JavaMan
Sep 19 '18 at 2:43
$begingroup$
The value of a function $f(x) $ may equal its limit $L$ as $xto a$. Check for example $f(x) =xsin (1/x)$ as $xto 0$. For a constant function the function value always equals its limits.
$endgroup$
– Paramanand Singh
Sep 19 '18 at 3:11
$begingroup$
I have seen this question in regard to the value of periodic number 0.(9) . In this case, indeed, some people try to say that it's not equal to 1 but the "limit" of this number is 1, disregarding the fact that the number is constant.
$endgroup$
– IMil
Sep 19 '18 at 5:19
1
1
$begingroup$
Limit of WHAT? There is no "input" to a constant. It is nonsense to write $lim_{x to x_0} pi$, for instance.
$endgroup$
– David G. Stork
Sep 19 '18 at 2:30
$begingroup$
Limit of WHAT? There is no "input" to a constant. It is nonsense to write $lim_{x to x_0} pi$, for instance.
$endgroup$
– David G. Stork
Sep 19 '18 at 2:30
15
15
$begingroup$
David I would interpret your limit to have value $pi$.
$endgroup$
– M_B
Sep 19 '18 at 2:35
$begingroup$
David I would interpret your limit to have value $pi$.
$endgroup$
– M_B
Sep 19 '18 at 2:35
15
15
$begingroup$
@DavidG.Stork It's not nonsense to write $lim_{x to x_0} pi$, but it is trivial. Here, $pi = f(x)$ is a constant function, so the definition is perfectly valid here.
$endgroup$
– JavaMan
Sep 19 '18 at 2:43
$begingroup$
@DavidG.Stork It's not nonsense to write $lim_{x to x_0} pi$, but it is trivial. Here, $pi = f(x)$ is a constant function, so the definition is perfectly valid here.
$endgroup$
– JavaMan
Sep 19 '18 at 2:43
$begingroup$
The value of a function $f(x) $ may equal its limit $L$ as $xto a$. Check for example $f(x) =xsin (1/x)$ as $xto 0$. For a constant function the function value always equals its limits.
$endgroup$
– Paramanand Singh
Sep 19 '18 at 3:11
$begingroup$
The value of a function $f(x) $ may equal its limit $L$ as $xto a$. Check for example $f(x) =xsin (1/x)$ as $xto 0$. For a constant function the function value always equals its limits.
$endgroup$
– Paramanand Singh
Sep 19 '18 at 3:11
$begingroup$
I have seen this question in regard to the value of periodic number 0.(9) . In this case, indeed, some people try to say that it's not equal to 1 but the "limit" of this number is 1, disregarding the fact that the number is constant.
$endgroup$
– IMil
Sep 19 '18 at 5:19
$begingroup$
I have seen this question in regard to the value of periodic number 0.(9) . In this case, indeed, some people try to say that it's not equal to 1 but the "limit" of this number is 1, disregarding the fact that the number is constant.
$endgroup$
– IMil
Sep 19 '18 at 5:19
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Your quote isn't a definition of a limit, but an English language description of what it computes.
If you look at the actual definition, such as the usual "epsilon-delta" definition, you'll see that it handles constant functions just fine, and in fact you have
$$ lim_{x to a} c = c $$
So, what has happened here is that there is a miscommunication — the meaning intended by the author of this wikipedia passage is not the meaning inferred by you the reader.
While you could decide that the author used sloppy language, or you could chalk up the whole thing to the imprecision of the English language, I think the following will be more useful:
- You need to refine your intuition about what "approach" means
Natural language tends to be exclusive of overly simple or degenerate cases — e.g. if you were to say "I live within 50 miles of Paris" in everyday conversation, the listener would assume that you don't live in Paris because it's expected that you would have said so.
Technical language tend to be more inclusive unless stated otherwise — e.g. that someone living in Paris does indeed live within 50 miles of Paris.
answered Sep 19 '18 at 3:14
HurkylHurkyl
111k9119262
$endgroup$
$begingroup$
Thank you, but how can I define "approach"?
$endgroup$
– Muhammad
Sep 19 '18 at 3:18
2
$begingroup$
@مُحَمَّدْ Look up $(epsilon-delta)$ definition. This is the usual definition used for defining a limit. You can find a review of this idea is here: khanacademy.org/math/calculus-all-old/… Temporarily forget the Wikipedia definition, and try to see if the idea of a limit makes sense under the "new" definition.
$endgroup$
– Dair
Sep 19 '18 at 3:23
2
$begingroup$
@مُحَمَّدْ Even without getting into technical definitions of limits, part of the problem is that you're confused about what is meant by "constant" in this context. What is meant isn't so much "constant" as "constant function". Although the number $2$ can't "approach" anything, a function which just happens to always be equal to $2$ certainly can.
$endgroup$
– Jack M
Sep 19 '18 at 9:06
$begingroup$
To the proposer: It can be very useful to think intuitively of a sequence as a series of events occurring in time. But eventually we need precise definitions. The $epsilon$-$delta$ formulation IS the whole definition of "approach". The great mathematician John von Neumann once said to a student, "Young man, in mathematics, one does not understand things One merely gets used to them."
$endgroup$
– DanielWainfleet
Sep 19 '18 at 10:40
add a comment |
$begingroup$
Let $f(x)=c$, where $c$ is some constant number, and suppose that the domain of $f$ is the real numbers. Then we can take the limit as $x$ approaches some value, say for example $$lim_{xrightarrow0}f(x).$$
Then by definition $f(x)=c hspace{0.1cm}$ for all $x$, so in particular $$lim_{xrightarrow0}f(x)=c.$$
Notice that the value of the function $f$ is constant, but the value of the independent variable $x$ changes.
answered Sep 19 '18 at 2:33
M_BM_B
415210
$endgroup$
$begingroup$
Okay, according to the quote "the input (or index)" here which is $x$ can approach but the function $f$ is still a constant that won't approach anything; so, it is not right to say that the function $f$ approaches $c$ as $x$ approaches $0$
$endgroup$
– Muhammad
Sep 19 '18 at 3:14
1
$begingroup$
It is still right to say f approaches c, even when f is constantly c. Draw a picture of the graph of $f(x)=3$. It is a horizontal line at height 3. Taking the limit as x approaches 0 is like sliding your finger towards 0 from the left or right along the graph. The x values get closer and closer to 0 and the f values “get closer and closer to 3”, even though in this case the f values are always 3.
$endgroup$
– M_B
Sep 19 '18 at 3:37
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@مُحَمَّدْ It doesn't need to approach -- it's already there!
$endgroup$
– David Richerby
Sep 19 '18 at 14:08
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David yes it is already there. Mathematically already being there is a special kind of ‘approaching’ where the distance to the target is always 0. This is how we use the word approach in math, even if it seems awkward when compared to your intuition about what it means.
$endgroup$
– M_B
Sep 19 '18 at 16:57
add a comment |
$begingroup$
First let's face the fact there may be more than one limit definition depending on the space in which you operate. For example a limit of a function for a given element of domain where both domain and codomain have some measure you'll likely go with the $epsilon - delta$ definition while if you're talking about a limit of an infinite sequence you need to have the sequence definition. They are related but not exactly the same.
The quote you cite from this Wikipedia article that you're referring to uses some textual explanation to make it easier to grasp the idea of a limit but it is not exact in
mathematical terms. This is supposed to only make it easier to understand the idea.
Now if you wanted to somehow define the "approach" meaning you'd have to look at the definition of the limes. Regardless of the chosen definition you have (rephrased to a bit more common English):
$c$ is a $lim$ of $f$ in (something - either an element in your domain or something that is next to your domain) if regardless how close you want to be to $c$ (by selecting an $epsilon > 0$ you can define some conditions narrowing which element $x$ you are allowed to pick from your domain - close to where you want to be in your domain so that for any $x$ from your domain selected in that way it is true that you are as close to the $c$ as you wanted to be, i.e.
$|f(x) - c| < epsilon$
This is not an actual quote, it's a generalization-rephrasing.
So still using a common English explanation if you get close in your domain to some $x_0$ you will be close to some $c$ in your codomain. If you are in $c$ already you are also very, very close.
Now there is nothing about approaching here, but in most cases most of your values even near your $x_0$ produce a value different than $c$ so using the term approach make it easier for the reader to understand. Yet if you're already in $c$ technically you don't approach it, but it doesn't invalidate the definition. So if you wan't to properly define approach you'd have to say you're either getting very close to the point or you are already there.
Note that your function may jump to $c$ sometimes and then get out of it. As long as you don't get away too far, you're still approaching $c$. But the common sense understanding of approach is different so you need to be careful not to confuse the precise mathematical and common sense meaning.
TL; DR
When trying to describe math in a common language, to make it understandable you use some simplification and loose actual math precision. You need to account that. Especially sources like Wikipedia should be treated with a grain of salt in such informal descriptions.
Terms written in script are not strict mathematical terms. Those terms are replaces with strict definitions/conditions but those definitions/conditions are sometimes difficult to grasp. I deliberately didn't want to use strict math terms.
answered Sep 19 '18 at 8:05
IsterIster
2216
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3 Answers
3
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3 Answers
3
active
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$begingroup$
Your quote isn't a definition of a limit, but an English language description of what it computes.
If you look at the actual definition, such as the usual "epsilon-delta" definition, you'll see that it handles constant functions just fine, and in fact you have
$$ lim_{x to a} c = c $$
So, what has happened here is that there is a miscommunication — the meaning intended by the author of this wikipedia passage is not the meaning inferred by you the reader.
While you could decide that the author used sloppy language, or you could chalk up the whole thing to the imprecision of the English language, I think the following will be more useful:
- You need to refine your intuition about what "approach" means
Natural language tends to be exclusive of overly simple or degenerate cases — e.g. if you were to say "I live within 50 miles of Paris" in everyday conversation, the listener would assume that you don't live in Paris because it's expected that you would have said so.
Technical language tend to be more inclusive unless stated otherwise — e.g. that someone living in Paris does indeed live within 50 miles of Paris.
answered Sep 19 '18 at 3:14
HurkylHurkyl
111k9119262
$endgroup$
$begingroup$
Thank you, but how can I define "approach"?
$endgroup$
– Muhammad
Sep 19 '18 at 3:18
2
$begingroup$
@مُحَمَّدْ Look up $(epsilon-delta)$ definition. This is the usual definition used for defining a limit. You can find a review of this idea is here: khanacademy.org/math/calculus-all-old/… Temporarily forget the Wikipedia definition, and try to see if the idea of a limit makes sense under the "new" definition.
$endgroup$
– Dair
Sep 19 '18 at 3:23
2
$begingroup$
@مُحَمَّدْ Even without getting into technical definitions of limits, part of the problem is that you're confused about what is meant by "constant" in this context. What is meant isn't so much "constant" as "constant function". Although the number $2$ can't "approach" anything, a function which just happens to always be equal to $2$ certainly can.
$endgroup$
– Jack M
Sep 19 '18 at 9:06
$begingroup$
To the proposer: It can be very useful to think intuitively of a sequence as a series of events occurring in time. But eventually we need precise definitions. The $epsilon$-$delta$ formulation IS the whole definition of "approach". The great mathematician John von Neumann once said to a student, "Young man, in mathematics, one does not understand things One merely gets used to them."
$endgroup$
– DanielWainfleet
Sep 19 '18 at 10:40
add a comment |
$begingroup$
Your quote isn't a definition of a limit, but an English language description of what it computes.
If you look at the actual definition, such as the usual "epsilon-delta" definition, you'll see that it handles constant functions just fine, and in fact you have
$$ lim_{x to a} c = c $$
So, what has happened here is that there is a miscommunication — the meaning intended by the author of this wikipedia passage is not the meaning inferred by you the reader.
While you could decide that the author used sloppy language, or you could chalk up the whole thing to the imprecision of the English language, I think the following will be more useful:
- You need to refine your intuition about what "approach" means
Natural language tends to be exclusive of overly simple or degenerate cases — e.g. if you were to say "I live within 50 miles of Paris" in everyday conversation, the listener would assume that you don't live in Paris because it's expected that you would have said so.
Technical language tend to be more inclusive unless stated otherwise — e.g. that someone living in Paris does indeed live within 50 miles of Paris.
answered Sep 19 '18 at 3:14
HurkylHurkyl
111k9119262
$endgroup$
$begingroup$
Thank you, but how can I define "approach"?
$endgroup$
– Muhammad
Sep 19 '18 at 3:18
2
$begingroup$
@مُحَمَّدْ Look up $(epsilon-delta)$ definition. This is the usual definition used for defining a limit. You can find a review of this idea is here: khanacademy.org/math/calculus-all-old/… Temporarily forget the Wikipedia definition, and try to see if the idea of a limit makes sense under the "new" definition.
$endgroup$
– Dair
Sep 19 '18 at 3:23
2
$begingroup$
@مُحَمَّدْ Even without getting into technical definitions of limits, part of the problem is that you're confused about what is meant by "constant" in this context. What is meant isn't so much "constant" as "constant function". Although the number $2$ can't "approach" anything, a function which just happens to always be equal to $2$ certainly can.
$endgroup$
– Jack M
Sep 19 '18 at 9:06
$begingroup$
To the proposer: It can be very useful to think intuitively of a sequence as a series of events occurring in time. But eventually we need precise definitions. The $epsilon$-$delta$ formulation IS the whole definition of "approach". The great mathematician John von Neumann once said to a student, "Young man, in mathematics, one does not understand things One merely gets used to them."
$endgroup$
– DanielWainfleet
Sep 19 '18 at 10:40
add a comment |
$begingroup$
Your quote isn't a definition of a limit, but an English language description of what it computes.
If you look at the actual definition, such as the usual "epsilon-delta" definition, you'll see that it handles constant functions just fine, and in fact you have
$$ lim_{x to a} c = c $$
So, what has happened here is that there is a miscommunication — the meaning intended by the author of this wikipedia passage is not the meaning inferred by you the reader.
While you could decide that the author used sloppy language, or you could chalk up the whole thing to the imprecision of the English language, I think the following will be more useful:
- You need to refine your intuition about what "approach" means
Natural language tends to be exclusive of overly simple or degenerate cases — e.g. if you were to say "I live within 50 miles of Paris" in everyday conversation, the listener would assume that you don't live in Paris because it's expected that you would have said so.
Technical language tend to be more inclusive unless stated otherwise — e.g. that someone living in Paris does indeed live within 50 miles of Paris.
answered Sep 19 '18 at 3:14
HurkylHurkyl
111k9119262
$endgroup$
Your quote isn't a definition of a limit, but an English language description of what it computes.
If you look at the actual definition, such as the usual "epsilon-delta" definition, you'll see that it handles constant functions just fine, and in fact you have
$$ lim_{x to a} c = c $$
So, what has happened here is that there is a miscommunication — the meaning intended by the author of this wikipedia passage is not the meaning inferred by you the reader.
While you could decide that the author used sloppy language, or you could chalk up the whole thing to the imprecision of the English language, I think the following will be more useful:
- You need to refine your intuition about what "approach" means
Natural language tends to be exclusive of overly simple or degenerate cases — e.g. if you were to say "I live within 50 miles of Paris" in everyday conversation, the listener would assume that you don't live in Paris because it's expected that you would have said so.
Technical language tend to be more inclusive unless stated otherwise — e.g. that someone living in Paris does indeed live within 50 miles of Paris.
answered Sep 19 '18 at 3:14
HurkylHurkyl
111k9119262
edited Sep 19 '18 at 3:19
answered Sep 19 '18 at 3:14
HurkylHurkyl
111k9119262
answered Sep 19 '18 at 3:14
HurkylHurkyl
111k9119262
answered Sep 19 '18 at 3:14
HurkylHurkyl
111k9119262
111k9119262
$begingroup$
Thank you, but how can I define "approach"?
$endgroup$
– Muhammad
Sep 19 '18 at 3:18
2
$begingroup$
@مُحَمَّدْ Look up $(epsilon-delta)$ definition. This is the usual definition used for defining a limit. You can find a review of this idea is here: khanacademy.org/math/calculus-all-old/… Temporarily forget the Wikipedia definition, and try to see if the idea of a limit makes sense under the "new" definition.
$endgroup$
– Dair
Sep 19 '18 at 3:23
2
$begingroup$
@مُحَمَّدْ Even without getting into technical definitions of limits, part of the problem is that you're confused about what is meant by "constant" in this context. What is meant isn't so much "constant" as "constant function". Although the number $2$ can't "approach" anything, a function which just happens to always be equal to $2$ certainly can.
$endgroup$
– Jack M
Sep 19 '18 at 9:06
$begingroup$
To the proposer: It can be very useful to think intuitively of a sequence as a series of events occurring in time. But eventually we need precise definitions. The $epsilon$-$delta$ formulation IS the whole definition of "approach". The great mathematician John von Neumann once said to a student, "Young man, in mathematics, one does not understand things One merely gets used to them."
$endgroup$
– DanielWainfleet
Sep 19 '18 at 10:40
add a comment |
$begingroup$
Thank you, but how can I define "approach"?
$endgroup$
– Muhammad
Sep 19 '18 at 3:18
2
$begingroup$
@مُحَمَّدْ Look up $(epsilon-delta)$ definition. This is the usual definition used for defining a limit. You can find a review of this idea is here: khanacademy.org/math/calculus-all-old/… Temporarily forget the Wikipedia definition, and try to see if the idea of a limit makes sense under the "new" definition.
$endgroup$
– Dair
Sep 19 '18 at 3:23
2
$begingroup$
@مُحَمَّدْ Even without getting into technical definitions of limits, part of the problem is that you're confused about what is meant by "constant" in this context. What is meant isn't so much "constant" as "constant function". Although the number $2$ can't "approach" anything, a function which just happens to always be equal to $2$ certainly can.
$endgroup$
– Jack M
Sep 19 '18 at 9:06
$begingroup$
To the proposer: It can be very useful to think intuitively of a sequence as a series of events occurring in time. But eventually we need precise definitions. The $epsilon$-$delta$ formulation IS the whole definition of "approach". The great mathematician John von Neumann once said to a student, "Young man, in mathematics, one does not understand things One merely gets used to them."
$endgroup$
– DanielWainfleet
Sep 19 '18 at 10:40
$begingroup$
Thank you, but how can I define "approach"?
$endgroup$
– Muhammad
Sep 19 '18 at 3:18
$begingroup$
Thank you, but how can I define "approach"?
$endgroup$
– Muhammad
Sep 19 '18 at 3:18
2
2
$begingroup$
@مُحَمَّدْ Look up $(epsilon-delta)$ definition. This is the usual definition used for defining a limit. You can find a review of this idea is here: khanacademy.org/math/calculus-all-old/… Temporarily forget the Wikipedia definition, and try to see if the idea of a limit makes sense under the "new" definition.
$endgroup$
– Dair
Sep 19 '18 at 3:23
$begingroup$
@مُحَمَّدْ Look up $(epsilon-delta)$ definition. This is the usual definition used for defining a limit. You can find a review of this idea is here: khanacademy.org/math/calculus-all-old/… Temporarily forget the Wikipedia definition, and try to see if the idea of a limit makes sense under the "new" definition.
$endgroup$
– Dair
Sep 19 '18 at 3:23
2
2
$begingroup$
@مُحَمَّدْ Even without getting into technical definitions of limits, part of the problem is that you're confused about what is meant by "constant" in this context. What is meant isn't so much "constant" as "constant function". Although the number $2$ can't "approach" anything, a function which just happens to always be equal to $2$ certainly can.
$endgroup$
– Jack M
Sep 19 '18 at 9:06
$begingroup$
@مُحَمَّدْ Even without getting into technical definitions of limits, part of the problem is that you're confused about what is meant by "constant" in this context. What is meant isn't so much "constant" as "constant function". Although the number $2$ can't "approach" anything, a function which just happens to always be equal to $2$ certainly can.
$endgroup$
– Jack M
Sep 19 '18 at 9:06
$begingroup$
To the proposer: It can be very useful to think intuitively of a sequence as a series of events occurring in time. But eventually we need precise definitions. The $epsilon$-$delta$ formulation IS the whole definition of "approach". The great mathematician John von Neumann once said to a student, "Young man, in mathematics, one does not understand things One merely gets used to them."
$endgroup$
– DanielWainfleet
Sep 19 '18 at 10:40
$begingroup$
To the proposer: It can be very useful to think intuitively of a sequence as a series of events occurring in time. But eventually we need precise definitions. The $epsilon$-$delta$ formulation IS the whole definition of "approach". The great mathematician John von Neumann once said to a student, "Young man, in mathematics, one does not understand things One merely gets used to them."
$endgroup$
– DanielWainfleet
Sep 19 '18 at 10:40
add a comment |
$begingroup$
Let $f(x)=c$, where $c$ is some constant number, and suppose that the domain of $f$ is the real numbers. Then we can take the limit as $x$ approaches some value, say for example $$lim_{xrightarrow0}f(x).$$
Then by definition $f(x)=c hspace{0.1cm}$ for all $x$, so in particular $$lim_{xrightarrow0}f(x)=c.$$
Notice that the value of the function $f$ is constant, but the value of the independent variable $x$ changes.
answered Sep 19 '18 at 2:33
M_BM_B
415210
$endgroup$
$begingroup$
Okay, according to the quote "the input (or index)" here which is $x$ can approach but the function $f$ is still a constant that won't approach anything; so, it is not right to say that the function $f$ approaches $c$ as $x$ approaches $0$
$endgroup$
– Muhammad
Sep 19 '18 at 3:14
1
$begingroup$
It is still right to say f approaches c, even when f is constantly c. Draw a picture of the graph of $f(x)=3$. It is a horizontal line at height 3. Taking the limit as x approaches 0 is like sliding your finger towards 0 from the left or right along the graph. The x values get closer and closer to 0 and the f values “get closer and closer to 3”, even though in this case the f values are always 3.
$endgroup$
– M_B
Sep 19 '18 at 3:37
$begingroup$
@مُحَمَّدْ It doesn't need to approach -- it's already there!
$endgroup$
– David Richerby
Sep 19 '18 at 14:08
$begingroup$
David yes it is already there. Mathematically already being there is a special kind of ‘approaching’ where the distance to the target is always 0. This is how we use the word approach in math, even if it seems awkward when compared to your intuition about what it means.
$endgroup$
– M_B
Sep 19 '18 at 16:57
add a comment |
$begingroup$
Let $f(x)=c$, where $c$ is some constant number, and suppose that the domain of $f$ is the real numbers. Then we can take the limit as $x$ approaches some value, say for example $$lim_{xrightarrow0}f(x).$$
Then by definition $f(x)=c hspace{0.1cm}$ for all $x$, so in particular $$lim_{xrightarrow0}f(x)=c.$$
Notice that the value of the function $f$ is constant, but the value of the independent variable $x$ changes.
answered Sep 19 '18 at 2:33
M_BM_B
415210
$endgroup$
$begingroup$
Okay, according to the quote "the input (or index)" here which is $x$ can approach but the function $f$ is still a constant that won't approach anything; so, it is not right to say that the function $f$ approaches $c$ as $x$ approaches $0$
$endgroup$
– Muhammad
Sep 19 '18 at 3:14
1
$begingroup$
It is still right to say f approaches c, even when f is constantly c. Draw a picture of the graph of $f(x)=3$. It is a horizontal line at height 3. Taking the limit as x approaches 0 is like sliding your finger towards 0 from the left or right along the graph. The x values get closer and closer to 0 and the f values “get closer and closer to 3”, even though in this case the f values are always 3.
$endgroup$
– M_B
Sep 19 '18 at 3:37
$begingroup$
@مُحَمَّدْ It doesn't need to approach -- it's already there!
$endgroup$
– David Richerby
Sep 19 '18 at 14:08
$begingroup$
David yes it is already there. Mathematically already being there is a special kind of ‘approaching’ where the distance to the target is always 0. This is how we use the word approach in math, even if it seems awkward when compared to your intuition about what it means.
$endgroup$
– M_B
Sep 19 '18 at 16:57
add a comment |
$begingroup$
Let $f(x)=c$, where $c$ is some constant number, and suppose that the domain of $f$ is the real numbers. Then we can take the limit as $x$ approaches some value, say for example $$lim_{xrightarrow0}f(x).$$
Then by definition $f(x)=c hspace{0.1cm}$ for all $x$, so in particular $$lim_{xrightarrow0}f(x)=c.$$
Notice that the value of the function $f$ is constant, but the value of the independent variable $x$ changes.
answered Sep 19 '18 at 2:33
M_BM_B
415210
$endgroup$
Let $f(x)=c$, where $c$ is some constant number, and suppose that the domain of $f$ is the real numbers. Then we can take the limit as $x$ approaches some value, say for example $$lim_{xrightarrow0}f(x).$$
Then by definition $f(x)=c hspace{0.1cm}$ for all $x$, so in particular $$lim_{xrightarrow0}f(x)=c.$$
Notice that the value of the function $f$ is constant, but the value of the independent variable $x$ changes.
answered Sep 19 '18 at 2:33
M_BM_B
415210
answered Sep 19 '18 at 2:33
M_BM_B
415210
answered Sep 19 '18 at 2:33
M_BM_B
415210
answered Sep 19 '18 at 2:33
M_BM_B
415210
415210
$begingroup$
Okay, according to the quote "the input (or index)" here which is $x$ can approach but the function $f$ is still a constant that won't approach anything; so, it is not right to say that the function $f$ approaches $c$ as $x$ approaches $0$
$endgroup$
– Muhammad
Sep 19 '18 at 3:14
1
$begingroup$
It is still right to say f approaches c, even when f is constantly c. Draw a picture of the graph of $f(x)=3$. It is a horizontal line at height 3. Taking the limit as x approaches 0 is like sliding your finger towards 0 from the left or right along the graph. The x values get closer and closer to 0 and the f values “get closer and closer to 3”, even though in this case the f values are always 3.
$endgroup$
– M_B
Sep 19 '18 at 3:37
$begingroup$
@مُحَمَّدْ It doesn't need to approach -- it's already there!
$endgroup$
– David Richerby
Sep 19 '18 at 14:08
$begingroup$
David yes it is already there. Mathematically already being there is a special kind of ‘approaching’ where the distance to the target is always 0. This is how we use the word approach in math, even if it seems awkward when compared to your intuition about what it means.
$endgroup$
– M_B
Sep 19 '18 at 16:57
add a comment |
$begingroup$
Okay, according to the quote "the input (or index)" here which is $x$ can approach but the function $f$ is still a constant that won't approach anything; so, it is not right to say that the function $f$ approaches $c$ as $x$ approaches $0$
$endgroup$
– Muhammad
Sep 19 '18 at 3:14
1
$begingroup$
It is still right to say f approaches c, even when f is constantly c. Draw a picture of the graph of $f(x)=3$. It is a horizontal line at height 3. Taking the limit as x approaches 0 is like sliding your finger towards 0 from the left or right along the graph. The x values get closer and closer to 0 and the f values “get closer and closer to 3”, even though in this case the f values are always 3.
$endgroup$
– M_B
Sep 19 '18 at 3:37
$begingroup$
@مُحَمَّدْ It doesn't need to approach -- it's already there!
$endgroup$
– David Richerby
Sep 19 '18 at 14:08
$begingroup$
David yes it is already there. Mathematically already being there is a special kind of ‘approaching’ where the distance to the target is always 0. This is how we use the word approach in math, even if it seems awkward when compared to your intuition about what it means.
$endgroup$
– M_B
Sep 19 '18 at 16:57
$begingroup$
Okay, according to the quote "the input (or index)" here which is $x$ can approach but the function $f$ is still a constant that won't approach anything; so, it is not right to say that the function $f$ approaches $c$ as $x$ approaches $0$
$endgroup$
– Muhammad
Sep 19 '18 at 3:14
$begingroup$
Okay, according to the quote "the input (or index)" here which is $x$ can approach but the function $f$ is still a constant that won't approach anything; so, it is not right to say that the function $f$ approaches $c$ as $x$ approaches $0$
$endgroup$
– Muhammad
Sep 19 '18 at 3:14
1
1
$begingroup$
It is still right to say f approaches c, even when f is constantly c. Draw a picture of the graph of $f(x)=3$. It is a horizontal line at height 3. Taking the limit as x approaches 0 is like sliding your finger towards 0 from the left or right along the graph. The x values get closer and closer to 0 and the f values “get closer and closer to 3”, even though in this case the f values are always 3.
$endgroup$
– M_B
Sep 19 '18 at 3:37
$begingroup$
It is still right to say f approaches c, even when f is constantly c. Draw a picture of the graph of $f(x)=3$. It is a horizontal line at height 3. Taking the limit as x approaches 0 is like sliding your finger towards 0 from the left or right along the graph. The x values get closer and closer to 0 and the f values “get closer and closer to 3”, even though in this case the f values are always 3.
$endgroup$
– M_B
Sep 19 '18 at 3:37
$begingroup$
@مُحَمَّدْ It doesn't need to approach -- it's already there!
$endgroup$
– David Richerby
Sep 19 '18 at 14:08
$begingroup$
@مُحَمَّدْ It doesn't need to approach -- it's already there!
$endgroup$
– David Richerby
Sep 19 '18 at 14:08
$begingroup$
David yes it is already there. Mathematically already being there is a special kind of ‘approaching’ where the distance to the target is always 0. This is how we use the word approach in math, even if it seems awkward when compared to your intuition about what it means.
$endgroup$
– M_B
Sep 19 '18 at 16:57
$begingroup$
David yes it is already there. Mathematically already being there is a special kind of ‘approaching’ where the distance to the target is always 0. This is how we use the word approach in math, even if it seems awkward when compared to your intuition about what it means.
$endgroup$
– M_B
Sep 19 '18 at 16:57
add a comment |
$begingroup$
First let's face the fact there may be more than one limit definition depending on the space in which you operate. For example a limit of a function for a given element of domain where both domain and codomain have some measure you'll likely go with the $epsilon - delta$ definition while if you're talking about a limit of an infinite sequence you need to have the sequence definition. They are related but not exactly the same.
The quote you cite from this Wikipedia article that you're referring to uses some textual explanation to make it easier to grasp the idea of a limit but it is not exact in
mathematical terms. This is supposed to only make it easier to understand the idea.
Now if you wanted to somehow define the "approach" meaning you'd have to look at the definition of the limes. Regardless of the chosen definition you have (rephrased to a bit more common English):
$c$ is a $lim$ of $f$ in (something - either an element in your domain or something that is next to your domain) if regardless how close you want to be to $c$ (by selecting an $epsilon > 0$ you can define some conditions narrowing which element $x$ you are allowed to pick from your domain - close to where you want to be in your domain so that for any $x$ from your domain selected in that way it is true that you are as close to the $c$ as you wanted to be, i.e.
$|f(x) - c| < epsilon$
This is not an actual quote, it's a generalization-rephrasing.
So still using a common English explanation if you get close in your domain to some $x_0$ you will be close to some $c$ in your codomain. If you are in $c$ already you are also very, very close.
Now there is nothing about approaching here, but in most cases most of your values even near your $x_0$ produce a value different than $c$ so using the term approach make it easier for the reader to understand. Yet if you're already in $c$ technically you don't approach it, but it doesn't invalidate the definition. So if you wan't to properly define approach you'd have to say you're either getting very close to the point or you are already there.
Note that your function may jump to $c$ sometimes and then get out of it. As long as you don't get away too far, you're still approaching $c$. But the common sense understanding of approach is different so you need to be careful not to confuse the precise mathematical and common sense meaning.
TL; DR
When trying to describe math in a common language, to make it understandable you use some simplification and loose actual math precision. You need to account that. Especially sources like Wikipedia should be treated with a grain of salt in such informal descriptions.
Terms written in script are not strict mathematical terms. Those terms are replaces with strict definitions/conditions but those definitions/conditions are sometimes difficult to grasp. I deliberately didn't want to use strict math terms.
answered Sep 19 '18 at 8:05
IsterIster
2216
$endgroup$
add a comment |
$begingroup$
First let's face the fact there may be more than one limit definition depending on the space in which you operate. For example a limit of a function for a given element of domain where both domain and codomain have some measure you'll likely go with the $epsilon - delta$ definition while if you're talking about a limit of an infinite sequence you need to have the sequence definition. They are related but not exactly the same.
The quote you cite from this Wikipedia article that you're referring to uses some textual explanation to make it easier to grasp the idea of a limit but it is not exact in
mathematical terms. This is supposed to only make it easier to understand the idea.
Now if you wanted to somehow define the "approach" meaning you'd have to look at the definition of the limes. Regardless of the chosen definition you have (rephrased to a bit more common English):
$c$ is a $lim$ of $f$ in (something - either an element in your domain or something that is next to your domain) if regardless how close you want to be to $c$ (by selecting an $epsilon > 0$ you can define some conditions narrowing which element $x$ you are allowed to pick from your domain - close to where you want to be in your domain so that for any $x$ from your domain selected in that way it is true that you are as close to the $c$ as you wanted to be, i.e.
$|f(x) - c| < epsilon$
This is not an actual quote, it's a generalization-rephrasing.
So still using a common English explanation if you get close in your domain to some $x_0$ you will be close to some $c$ in your codomain. If you are in $c$ already you are also very, very close.
Now there is nothing about approaching here, but in most cases most of your values even near your $x_0$ produce a value different than $c$ so using the term approach make it easier for the reader to understand. Yet if you're already in $c$ technically you don't approach it, but it doesn't invalidate the definition. So if you wan't to properly define approach you'd have to say you're either getting very close to the point or you are already there.
Note that your function may jump to $c$ sometimes and then get out of it. As long as you don't get away too far, you're still approaching $c$. But the common sense understanding of approach is different so you need to be careful not to confuse the precise mathematical and common sense meaning.
TL; DR
When trying to describe math in a common language, to make it understandable you use some simplification and loose actual math precision. You need to account that. Especially sources like Wikipedia should be treated with a grain of salt in such informal descriptions.
Terms written in script are not strict mathematical terms. Those terms are replaces with strict definitions/conditions but those definitions/conditions are sometimes difficult to grasp. I deliberately didn't want to use strict math terms.
answered Sep 19 '18 at 8:05
IsterIster
2216
$endgroup$
add a comment |
$begingroup$
First let's face the fact there may be more than one limit definition depending on the space in which you operate. For example a limit of a function for a given element of domain where both domain and codomain have some measure you'll likely go with the $epsilon - delta$ definition while if you're talking about a limit of an infinite sequence you need to have the sequence definition. They are related but not exactly the same.
The quote you cite from this Wikipedia article that you're referring to uses some textual explanation to make it easier to grasp the idea of a limit but it is not exact in
mathematical terms. This is supposed to only make it easier to understand the idea.
Now if you wanted to somehow define the "approach" meaning you'd have to look at the definition of the limes. Regardless of the chosen definition you have (rephrased to a bit more common English):
$c$ is a $lim$ of $f$ in (something - either an element in your domain or something that is next to your domain) if regardless how close you want to be to $c$ (by selecting an $epsilon > 0$ you can define some conditions narrowing which element $x$ you are allowed to pick from your domain - close to where you want to be in your domain so that for any $x$ from your domain selected in that way it is true that you are as close to the $c$ as you wanted to be, i.e.
$|f(x) - c| < epsilon$
This is not an actual quote, it's a generalization-rephrasing.
So still using a common English explanation if you get close in your domain to some $x_0$ you will be close to some $c$ in your codomain. If you are in $c$ already you are also very, very close.
Now there is nothing about approaching here, but in most cases most of your values even near your $x_0$ produce a value different than $c$ so using the term approach make it easier for the reader to understand. Yet if you're already in $c$ technically you don't approach it, but it doesn't invalidate the definition. So if you wan't to properly define approach you'd have to say you're either getting very close to the point or you are already there.
Note that your function may jump to $c$ sometimes and then get out of it. As long as you don't get away too far, you're still approaching $c$. But the common sense understanding of approach is different so you need to be careful not to confuse the precise mathematical and common sense meaning.
TL; DR
When trying to describe math in a common language, to make it understandable you use some simplification and loose actual math precision. You need to account that. Especially sources like Wikipedia should be treated with a grain of salt in such informal descriptions.
Terms written in script are not strict mathematical terms. Those terms are replaces with strict definitions/conditions but those definitions/conditions are sometimes difficult to grasp. I deliberately didn't want to use strict math terms.
answered Sep 19 '18 at 8:05
IsterIster
2216
$endgroup$
First let's face the fact there may be more than one limit definition depending on the space in which you operate. For example a limit of a function for a given element of domain where both domain and codomain have some measure you'll likely go with the $epsilon - delta$ definition while if you're talking about a limit of an infinite sequence you need to have the sequence definition. They are related but not exactly the same.
The quote you cite from this Wikipedia article that you're referring to uses some textual explanation to make it easier to grasp the idea of a limit but it is not exact in
mathematical terms. This is supposed to only make it easier to understand the idea.
Now if you wanted to somehow define the "approach" meaning you'd have to look at the definition of the limes. Regardless of the chosen definition you have (rephrased to a bit more common English):
$c$ is a $lim$ of $f$ in (something - either an element in your domain or something that is next to your domain) if regardless how close you want to be to $c$ (by selecting an $epsilon > 0$ you can define some conditions narrowing which element $x$ you are allowed to pick from your domain - close to where you want to be in your domain so that for any $x$ from your domain selected in that way it is true that you are as close to the $c$ as you wanted to be, i.e.
$|f(x) - c| < epsilon$
This is not an actual quote, it's a generalization-rephrasing.
So still using a common English explanation if you get close in your domain to some $x_0$ you will be close to some $c$ in your codomain. If you are in $c$ already you are also very, very close.
Now there is nothing about approaching here, but in most cases most of your values even near your $x_0$ produce a value different than $c$ so using the term approach make it easier for the reader to understand. Yet if you're already in $c$ technically you don't approach it, but it doesn't invalidate the definition. So if you wan't to properly define approach you'd have to say you're either getting very close to the point or you are already there.
Note that your function may jump to $c$ sometimes and then get out of it. As long as you don't get away too far, you're still approaching $c$. But the common sense understanding of approach is different so you need to be careful not to confuse the precise mathematical and common sense meaning.
TL; DR
When trying to describe math in a common language, to make it understandable you use some simplification and loose actual math precision. You need to account that. Especially sources like Wikipedia should be treated with a grain of salt in such informal descriptions.
Terms written in script are not strict mathematical terms. Those terms are replaces with strict definitions/conditions but those definitions/conditions are sometimes difficult to grasp. I deliberately didn't want to use strict math terms.
answered Sep 19 '18 at 8:05
IsterIster
2216
edited Sep 19 '18 at 8:10
answered Sep 19 '18 at 8:05
IsterIster
2216
answered Sep 19 '18 at 8:05
IsterIster
2216
answered Sep 19 '18 at 8:05
IsterIster
2216
2216
add a comment |
add a comment |
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1
$begingroup$
Limit of WHAT? There is no "input" to a constant. It is nonsense to write $lim_{x to x_0} pi$, for instance.
$endgroup$
– David G. Stork
Sep 19 '18 at 2:30
15
$begingroup$
David I would interpret your limit to have value $pi$.
$endgroup$
– M_B
Sep 19 '18 at 2:35
15
$begingroup$
@DavidG.Stork It's not nonsense to write $lim_{x to x_0} pi$, but it is trivial. Here, $pi = f(x)$ is a constant function, so the definition is perfectly valid here.
$endgroup$
– JavaMan
Sep 19 '18 at 2:43
$begingroup$
The value of a function $f(x) $ may equal its limit $L$ as $xto a$. Check for example $f(x) =xsin (1/x)$ as $xto 0$. For a constant function the function value always equals its limits.
$endgroup$
– Paramanand Singh
Sep 19 '18 at 3:11
$begingroup$
I have seen this question in regard to the value of periodic number 0.(9) . In this case, indeed, some people try to say that it's not equal to 1 but the "limit" of this number is 1, disregarding the fact that the number is constant.
$endgroup$
– IMil
Sep 19 '18 at 5:19