Why can I not put the limit inside the limit definition of $e$?Proof that the solution to cosx = x, is the...
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Why can I not put the limit inside the limit definition of $e$?
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$begingroup$
We know $e:= lim_{nto infty}(1+frac{1}{n})^n$
We also know $x^n$ is continuous. Why is there a contradiction in the following?
$e=lim_{nto infty}(1+frac{1}{n})^n=(lim_{nto infty}(1+frac{1}{n}))^n=1$
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 19 mins ago
YuiTo Cheng
3,25371245
asked 6 hours ago
abcdabcd
41
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 $e:= lim_{nto infty}(1+frac{1}{n})^n$
We also know $x^n$ is continuous. Why is there a contradiction in the following?
$e=lim_{nto infty}(1+frac{1}{n})^n=(lim_{nto infty}(1+frac{1}{n}))^n=1$
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 19 mins ago
YuiTo Cheng
3,25371245
asked 6 hours ago
abcdabcd
41
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.
$endgroup$
3
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
6 hours ago
2
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
6 hours ago
2
$begingroup$
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
4 hours ago
add a comment |
$begingroup$
We know $e:= lim_{nto infty}(1+frac{1}{n})^n$
We also know $x^n$ is continuous. Why is there a contradiction in the following?
$e=lim_{nto infty}(1+frac{1}{n})^n=(lim_{nto infty}(1+frac{1}{n}))^n=1$
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 19 mins ago
YuiTo Cheng
3,25371245
asked 6 hours ago
abcdabcd
41
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 $e:= lim_{nto infty}(1+frac{1}{n})^n$
We also know $x^n$ is continuous. Why is there a contradiction in the following?
$e=lim_{nto infty}(1+frac{1}{n})^n=(lim_{nto infty}(1+frac{1}{n}))^n=1$
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 19 mins ago
YuiTo Cheng
3,25371245
asked 6 hours ago
abcdabcd
41
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 19 mins ago
YuiTo Cheng
3,25371245
asked 6 hours ago
abcdabcd
41
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 19 mins ago
YuiTo Cheng
3,25371245
edited 19 mins ago
YuiTo Cheng
3,25371245
edited 19 mins ago
YuiTo Cheng
3,25371245
3,25371245
asked 6 hours ago
abcdabcd
41
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 6 hours ago
abcdabcd
41
asked 6 hours ago
abcdabcd
41
41
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.
3
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
6 hours ago
2
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
6 hours ago
2
$begingroup$
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
4 hours ago
add a comment |
3
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
6 hours ago
2
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
6 hours ago
2
$begingroup$
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
4 hours ago
3
3
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
6 hours ago
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
6 hours ago
2
2
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
6 hours ago
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
6 hours ago
2
2
$begingroup$
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
4 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
4 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 6 hours ago
PawelPawel
3,3341022
$endgroup$
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
2 hours ago
add a comment |
$begingroup$
The actual problem is you can't write $(lim_{ntoinfty}f(n))^n$, 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 $(lim_{xto 2}x)+x$.
answered 6 hours ago
J.G.J.G.
37.7k23656
$endgroup$
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 6 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2761625
$endgroup$
add a comment |
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3 Answers
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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 6 hours ago
PawelPawel
3,3341022
$endgroup$
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
2 hours 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 6 hours ago
PawelPawel
3,3341022
$endgroup$
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
2 hours 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 6 hours ago
PawelPawel
3,3341022
$endgroup$
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 6 hours ago
PawelPawel
3,3341022
answered 6 hours ago
PawelPawel
3,3341022
answered 6 hours ago
PawelPawel
3,3341022
answered 6 hours ago
PawelPawel
3,3341022
3,3341022
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
2 hours ago
add a comment |
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
2 hours ago
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
2 hours ago
$begingroup$
Well explained. The limit operation and the variable bound to it go together.
$endgroup$
– Paramanand Singh
2 hours ago
add a comment |
$begingroup$
The actual problem is you can't write $(lim_{ntoinfty}f(n))^n$, 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 $(lim_{xto 2}x)+x$.
answered 6 hours ago
J.G.J.G.
37.7k23656
$endgroup$
add a comment |
$begingroup$
The actual problem is you can't write $(lim_{ntoinfty}f(n))^n$, 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 $(lim_{xto 2}x)+x$.
answered 6 hours ago
J.G.J.G.
37.7k23656
$endgroup$
add a comment |
$begingroup$
The actual problem is you can't write $(lim_{ntoinfty}f(n))^n$, 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 $(lim_{xto 2}x)+x$.
answered 6 hours ago
J.G.J.G.
37.7k23656
$endgroup$
The actual problem is you can't write $(lim_{ntoinfty}f(n))^n$, 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 $(lim_{xto 2}x)+x$.
answered 6 hours ago
J.G.J.G.
37.7k23656
answered 6 hours ago
J.G.J.G.
37.7k23656
answered 6 hours ago
J.G.J.G.
37.7k23656
answered 6 hours ago
J.G.J.G.
37.7k23656
37.7k23656
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 6 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2761625
$endgroup$
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 6 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2761625
$endgroup$
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 6 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2761625
$endgroup$
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 6 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2761625
answered 6 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2761625
answered 6 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2761625
answered 6 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,2761625
4,2761625
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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3
$begingroup$
Because $n$ isn't a fixed number, so it's presence outside of the limit is meaningless
$endgroup$
– MPW
6 hours ago
2
$begingroup$
We say $1^infty$ is an indeterminete form. Look that up in your calculus textbook.
$endgroup$
– GEdgar
6 hours ago
2
$begingroup$
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
4 hours ago