Why can I not put the limit inside the limit definition of e?Proof that the solution to cosx = x, is the limit of a recursive sequence.Move derivative inside limit.Showing a limit does not exist using Cauchy's $epsilon, delta $ limit definitionUniformly continuous function which is integrable but does not have a limitWhy is the limit not $ln(x)$?Limit definition of integrationComputing limit using definition of eThe Limit Comparison Test V1What do I miss on this function? $f(t) = (t-1)^s/2-t^s/2+1$Determining if the limit exists of the sequence
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Why can I not put the limit inside the limit definition of e?
Would it be possible to set up a franchise in the ancient world?
Character had a different name in the past. Which name should I use in a flashback?
Why can I not put the limit inside the limit definition of e?
Proof that the solution to cosx = x, is the limit of a recursive sequence.Move derivative inside limit.Showing a limit does not exist using Cauchy's $epsilon, delta $ limit definitionUniformly continuous function which is integrable but does not have a limitWhy is the limit not $ln(x)$?Limit definition of integrationComputing limit using definition of eThe Limit Comparison Test V1What do I miss on this function? $f(t) = (t-1)^s/2-t^s/2+1$Determining if the limit exists of the sequence
$begingroup$
We know $e:= lim_nto infty(1+frac1n)^n$
We also know $x^n$ is continuous. Why is there a contradiction in the following ?
$e=lim_nto infty(1+frac1n)^n=(lim_nto infty(1+frac1n))^n=1$
Is it because $x^n$ isn't really a continuous function, because the $n$ is not fixed but rather something like infinity ?
real-analysis calculus
edited 5 hours ago
Bernard
126k743120
asked 5 hours ago
abcdabcd
61
New contributor
abcd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
We know $e:= lim_nto infty(1+frac1n)^n$
We also know $x^n$ is continuous. Why is there a contradiction in the following ?
$e=lim_nto infty(1+frac1n)^n=(lim_nto infty(1+frac1n))^n=1$
Is it because $x^n$ isn't really a continuous function, because the $n$ is not fixed but rather something like infinity ?
real-analysis calculus
edited 5 hours ago
Bernard
126k743120
asked 5 hours ago
abcdabcd
61
New contributor
abcd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
5 hours ago
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
5 hours ago
1
$begingroup$
Note that once you put "the limit inside the limit definition" as shown, the exponent $n$ is no longer in the scope of taking a limit as $ntoinfty$. As such, the meaning of $n$ the exponent is no longer tied to the $n$ that is in the innermost parentheses. Both need to be taken to the limit together to have the original meaning.
$endgroup$
– hardmath
3 hours ago
add a comment |
$begingroup$
We know $e:= lim_nto infty(1+frac1n)^n$
We also know $x^n$ is continuous. Why is there a contradiction in the following ?
$e=lim_nto infty(1+frac1n)^n=(lim_nto infty(1+frac1n))^n=1$
Is it because $x^n$ isn't really a continuous function, because the $n$ is not fixed but rather something like infinity ?
real-analysis calculus
edited 5 hours ago
Bernard
126k743120
asked 5 hours ago
abcdabcd
61
New contributor
abcd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We know $e:= lim_nto infty(1+frac1n)^n$
We also know $x^n$ is continuous. Why is there a contradiction in the following ?
$e=lim_nto infty(1+frac1n)^n=(lim_nto infty(1+frac1n))^n=1$
Is it because $x^n$ isn't really a continuous function, because the $n$ is not fixed but rather something like infinity ?
real-analysis calculus
real-analysis calculus
edited 5 hours ago
Bernard
126k743120
asked 5 hours ago
abcdabcd
61
New contributor
abcd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Bernard
126k743120
asked 5 hours ago
abcdabcd
61
New contributor
abcd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Bernard
126k743120
edited 5 hours ago
Bernard
126k743120
edited 5 hours ago
Bernard
126k743120
126k743120
asked 5 hours ago
abcdabcd
61
New contributor
abcd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
abcdabcd
61
asked 5 hours ago
abcdabcd
61
61
New contributor
abcd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
abcd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
5 hours ago
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
5 hours ago
1
$begingroup$
Note that once you put "the limit inside the limit definition" as shown, the exponent $n$ is no longer in the scope of taking a limit as $ntoinfty$. As such, the meaning of $n$ the exponent is no longer tied to the $n$ that is in the innermost parentheses. Both need to be taken to the limit together to have the original meaning.
$endgroup$
– hardmath
3 hours ago
add a comment |
2
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
5 hours ago
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
5 hours ago
1
$begingroup$
Note that once you put "the limit inside the limit definition" as shown, the exponent $n$ is no longer in the scope of taking a limit as $ntoinfty$. As such, the meaning of $n$ the exponent is no longer tied to the $n$ that is in the innermost parentheses. Both need to be taken to the limit together to have the original meaning.
$endgroup$
– hardmath
3 hours ago
2
2
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
5 hours ago
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
5 hours ago
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
5 hours ago
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
5 hours ago
1
1
$begingroup$
Note that once you put "the limit inside the limit definition" as shown, the exponent $n$ is no longer in the scope of taking a limit as $ntoinfty$. As such, the meaning of $n$ the exponent is no longer tied to the $n$ that is in the innermost parentheses. Both need to be taken to the limit together to have the original meaning.
$endgroup$
– hardmath
3 hours ago
$begingroup$
Note that once you put "the limit inside the limit definition" as shown, the exponent $n$ is no longer in the scope of taking a limit as $ntoinfty$. As such, the meaning of $n$ the exponent is no longer tied to the $n$ that is in the innermost parentheses. Both need to be taken to the limit together to have the original meaning.
$endgroup$
– hardmath
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can't move a limit that depends on $n$ "inside" the function if the outside function depends on $n$. For example, it is clearly true that
$$lim_n to infty 2^n= infty .$$
However, if you move the limit "inside," you get
$$lim_n to infty 2^n= (lim_n to infty 2)^n=2^n$$
which does not make much sense, since you do not know what $n$ is... $n$ was supposed to go to infinity. However, since you "got rid of" the limit, $n$ has no meaning anymore.
In other words, whenever you are calculating limits, $n$ should disappear at the same step as the words $lim_n to infty$. If these two things don't go away at the same time, most likely it is a sign that you did something wrong.
I hope that helps!
answered 5 hours ago
PawelPawel
3,3241022
$endgroup$
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
1 hour ago
add a comment |
$begingroup$
The actual problem is you can't write $(lim_ntoinftyf(n))^n$, or anything else that uses $n$ outside the limit, because it's a dummy variable that exists only inside the limit. The role of $infty$ is irrelevant; you also can't, for the same reason, jump from $lim_xto 2(x+x)=4$ to writing $(lim_xto 2x)+x$.
answered 5 hours ago
J.G.J.G.
37.6k23656
$endgroup$
add a comment |
$begingroup$
Because the exponent is variable depending. In this case you get $1^infty$ which is indeterminate, or means merely that this limit can be anything.
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2661625
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can't move a limit that depends on $n$ "inside" the function if the outside function depends on $n$. For example, it is clearly true that
$$lim_n to infty 2^n= infty .$$
However, if you move the limit "inside," you get
$$lim_n to infty 2^n= (lim_n to infty 2)^n=2^n$$
which does not make much sense, since you do not know what $n$ is... $n$ was supposed to go to infinity. However, since you "got rid of" the limit, $n$ has no meaning anymore.
In other words, whenever you are calculating limits, $n$ should disappear at the same step as the words $lim_n to infty$. If these two things don't go away at the same time, most likely it is a sign that you did something wrong.
I hope that helps!
answered 5 hours ago
PawelPawel
3,3241022
$endgroup$
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
1 hour ago
add a comment |
$begingroup$
You can't move a limit that depends on $n$ "inside" the function if the outside function depends on $n$. For example, it is clearly true that
$$lim_n to infty 2^n= infty .$$
However, if you move the limit "inside," you get
$$lim_n to infty 2^n= (lim_n to infty 2)^n=2^n$$
which does not make much sense, since you do not know what $n$ is... $n$ was supposed to go to infinity. However, since you "got rid of" the limit, $n$ has no meaning anymore.
In other words, whenever you are calculating limits, $n$ should disappear at the same step as the words $lim_n to infty$. If these two things don't go away at the same time, most likely it is a sign that you did something wrong.
I hope that helps!
answered 5 hours ago
PawelPawel
3,3241022
$endgroup$
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
1 hour ago
add a comment |
$begingroup$
You can't move a limit that depends on $n$ "inside" the function if the outside function depends on $n$. For example, it is clearly true that
$$lim_n to infty 2^n= infty .$$
However, if you move the limit "inside," you get
$$lim_n to infty 2^n= (lim_n to infty 2)^n=2^n$$
which does not make much sense, since you do not know what $n$ is... $n$ was supposed to go to infinity. However, since you "got rid of" the limit, $n$ has no meaning anymore.
In other words, whenever you are calculating limits, $n$ should disappear at the same step as the words $lim_n to infty$. If these two things don't go away at the same time, most likely it is a sign that you did something wrong.
I hope that helps!
answered 5 hours ago
PawelPawel
3,3241022
$endgroup$
You can't move a limit that depends on $n$ "inside" the function if the outside function depends on $n$. For example, it is clearly true that
$$lim_n to infty 2^n= infty .$$
However, if you move the limit "inside," you get
$$lim_n to infty 2^n= (lim_n to infty 2)^n=2^n$$
which does not make much sense, since you do not know what $n$ is... $n$ was supposed to go to infinity. However, since you "got rid of" the limit, $n$ has no meaning anymore.
In other words, whenever you are calculating limits, $n$ should disappear at the same step as the words $lim_n to infty$. If these two things don't go away at the same time, most likely it is a sign that you did something wrong.
I hope that helps!
answered 5 hours ago
PawelPawel
3,3241022
answered 5 hours ago
PawelPawel
3,3241022
answered 5 hours ago
PawelPawel
3,3241022
answered 5 hours ago
PawelPawel
3,3241022
3,3241022
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
1 hour ago
add a comment |
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
1 hour ago
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
1 hour ago
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
1 hour ago
add a comment |
$begingroup$
The actual problem is you can't write $(lim_ntoinftyf(n))^n$, or anything else that uses $n$ outside the limit, because it's a dummy variable that exists only inside the limit. The role of $infty$ is irrelevant; you also can't, for the same reason, jump from $lim_xto 2(x+x)=4$ to writing $(lim_xto 2x)+x$.
answered 5 hours ago
J.G.J.G.
37.6k23656
$endgroup$
add a comment |
$begingroup$
The actual problem is you can't write $(lim_ntoinftyf(n))^n$, or anything else that uses $n$ outside the limit, because it's a dummy variable that exists only inside the limit. The role of $infty$ is irrelevant; you also can't, for the same reason, jump from $lim_xto 2(x+x)=4$ to writing $(lim_xto 2x)+x$.
answered 5 hours ago
J.G.J.G.
37.6k23656
$endgroup$
add a comment |
$begingroup$
The actual problem is you can't write $(lim_ntoinftyf(n))^n$, or anything else that uses $n$ outside the limit, because it's a dummy variable that exists only inside the limit. The role of $infty$ is irrelevant; you also can't, for the same reason, jump from $lim_xto 2(x+x)=4$ to writing $(lim_xto 2x)+x$.
answered 5 hours ago
J.G.J.G.
37.6k23656
$endgroup$
The actual problem is you can't write $(lim_ntoinftyf(n))^n$, or anything else that uses $n$ outside the limit, because it's a dummy variable that exists only inside the limit. The role of $infty$ is irrelevant; you also can't, for the same reason, jump from $lim_xto 2(x+x)=4$ to writing $(lim_xto 2x)+x$.
answered 5 hours ago
J.G.J.G.
37.6k23656
answered 5 hours ago
J.G.J.G.
37.6k23656
answered 5 hours ago
J.G.J.G.
37.6k23656
answered 5 hours ago
J.G.J.G.
37.6k23656
37.6k23656
add a comment |
add a comment |
$begingroup$
Because the exponent is variable depending. In this case you get $1^infty$ which is indeterminate, or means merely that this limit can be anything.
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2661625
$endgroup$
add a comment |
$begingroup$
Because the exponent is variable depending. In this case you get $1^infty$ which is indeterminate, or means merely that this limit can be anything.
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2661625
$endgroup$
add a comment |
$begingroup$
Because the exponent is variable depending. In this case you get $1^infty$ which is indeterminate, or means merely that this limit can be anything.
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2661625
$endgroup$
Because the exponent is variable depending. In this case you get $1^infty$ which is indeterminate, or means merely that this limit can be anything.
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2661625
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2661625
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2661625
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2661625
4,2661625
add a comment |
add a comment |
abcd is a new contributor. Be nice, and check out our Code of Conduct.
abcd is a new contributor. Be nice, and check out our Code of Conduct.
abcd is a new contributor. Be nice, and check out our Code of Conduct.
abcd is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
5 hours ago
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
5 hours ago
1
$begingroup$
Note that once you put "the limit inside the limit definition" as shown, the exponent $n$ is no longer in the scope of taking a limit as $ntoinfty$. As such, the meaning of $n$ the exponent is no longer tied to the $n$ that is in the innermost parentheses. Both need to be taken to the limit together to have the original meaning.
$endgroup$
– hardmath
3 hours ago