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Limit of $lim_ntoinfty n-neleft(1-frac1nright)^n$? [closed]
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How do I calculate
$$lim_ntoinfty n-neleft(1-frac1nright)^n$$
Edit: changed it to correct question with $1-1/n$
calculus limits
edited Jul 31 at 2:28
user 108128
18.9k41544
asked Jul 31 at 1:43
MM TT
385
closed as off-topic by user 108128, Paramanand Singh, Arnaud D., John Ma, José Carlos Santos Jul 31 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser 108128, Paramanand Singh, Arnaud D., John Ma, Jos̩ Carlos Santos
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up vote
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down vote
favorite
How do I calculate
$$lim_ntoinfty n-neleft(1-frac1nright)^n$$
Edit: changed it to correct question with $1-1/n$
calculus limits
edited Jul 31 at 2:28
user 108128
18.9k41544
asked Jul 31 at 1:43
MM TT
385
closed as off-topic by user 108128, Paramanand Singh, Arnaud D., John Ma, José Carlos Santos Jul 31 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser 108128, Paramanand Singh, Arnaud D., John Ma, Jos̩ Carlos Santos
1
Write it as $n(1-e(1+1/n)^n)$, use that $lim_ntoinfty(1+1/n)^n=e$. Therefore $1-e(1+1/n)^nto1-e^2<0$. Finally, since $ntoinfty$, then the given limit tends to $-infty$.
– spiralstotheleft
Jul 31 at 1:46
1
For the new one you can write it as $frac1-e^fracln(1-1/n)1/n+11/n$. Compute a few terms of the series expansion (in powers of $1/n$) of $fracln(1-1/n)1/n+1$ in powers of $1/n$ to get $-1/(2 n) - 1/(3 n^2)+...$. Then compose with the series of $e^x$ to get the initial terms of the series (in powers of $1/n$) of the original expression $1/2 + 5/(24 n)+...$. Therefore, the limit is $1/2$.
– spiralstotheleft
Jul 31 at 2:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How do I calculate
$$lim_ntoinfty n-neleft(1-frac1nright)^n$$
Edit: changed it to correct question with $1-1/n$
calculus limits
edited Jul 31 at 2:28
user 108128
18.9k41544
asked Jul 31 at 1:43
MM TT
385
How do I calculate
$$lim_ntoinfty n-neleft(1-frac1nright)^n$$
Edit: changed it to correct question with $1-1/n$
calculus limits
edited Jul 31 at 2:28
user 108128
18.9k41544
asked Jul 31 at 1:43
MM TT
385
edited Jul 31 at 2:28
user 108128
18.9k41544
edited Jul 31 at 2:28
user 108128
18.9k41544
edited Jul 31 at 2:28
user 108128
18.9k41544
18.9k41544
asked Jul 31 at 1:43
MM TT
385
asked Jul 31 at 1:43
MM TT
385
asked Jul 31 at 1:43
MM TT
385
385
closed as off-topic by user 108128, Paramanand Singh, Arnaud D., John Ma, José Carlos Santos Jul 31 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser 108128, Paramanand Singh, Arnaud D., John Ma, Jos̩ Carlos Santos
closed as off-topic by user 108128, Paramanand Singh, Arnaud D., John Ma, José Carlos Santos Jul 31 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." Рuser 108128, Paramanand Singh, Arnaud D., John Ma, Jos̩ Carlos Santos
1
Write it as $n(1-e(1+1/n)^n)$, use that $lim_ntoinfty(1+1/n)^n=e$. Therefore $1-e(1+1/n)^nto1-e^2<0$. Finally, since $ntoinfty$, then the given limit tends to $-infty$.
– spiralstotheleft
Jul 31 at 1:46
1
For the new one you can write it as $frac1-e^fracln(1-1/n)1/n+11/n$. Compute a few terms of the series expansion (in powers of $1/n$) of $fracln(1-1/n)1/n+1$ in powers of $1/n$ to get $-1/(2 n) - 1/(3 n^2)+...$. Then compose with the series of $e^x$ to get the initial terms of the series (in powers of $1/n$) of the original expression $1/2 + 5/(24 n)+...$. Therefore, the limit is $1/2$.
– spiralstotheleft
Jul 31 at 2:06
add a comment |Â
1
Write it as $n(1-e(1+1/n)^n)$, use that $lim_ntoinfty(1+1/n)^n=e$. Therefore $1-e(1+1/n)^nto1-e^2<0$. Finally, since $ntoinfty$, then the given limit tends to $-infty$.
– spiralstotheleft
Jul 31 at 1:46
1
For the new one you can write it as $frac1-e^fracln(1-1/n)1/n+11/n$. Compute a few terms of the series expansion (in powers of $1/n$) of $fracln(1-1/n)1/n+1$ in powers of $1/n$ to get $-1/(2 n) - 1/(3 n^2)+...$. Then compose with the series of $e^x$ to get the initial terms of the series (in powers of $1/n$) of the original expression $1/2 + 5/(24 n)+...$. Therefore, the limit is $1/2$.
– spiralstotheleft
Jul 31 at 2:06
1
1
Write it as $n(1-e(1+1/n)^n)$, use that $lim_ntoinfty(1+1/n)^n=e$. Therefore $1-e(1+1/n)^nto1-e^2<0$. Finally, since $ntoinfty$, then the given limit tends to $-infty$.
– spiralstotheleft
Jul 31 at 1:46
Write it as $n(1-e(1+1/n)^n)$, use that $lim_ntoinfty(1+1/n)^n=e$. Therefore $1-e(1+1/n)^nto1-e^2<0$. Finally, since $ntoinfty$, then the given limit tends to $-infty$.
– spiralstotheleft
Jul 31 at 1:46
1
1
For the new one you can write it as $frac1-e^fracln(1-1/n)1/n+11/n$. Compute a few terms of the series expansion (in powers of $1/n$) of $fracln(1-1/n)1/n+1$ in powers of $1/n$ to get $-1/(2 n) - 1/(3 n^2)+...$. Then compose with the series of $e^x$ to get the initial terms of the series (in powers of $1/n$) of the original expression $1/2 + 5/(24 n)+...$. Therefore, the limit is $1/2$.
– spiralstotheleft
Jul 31 at 2:06
For the new one you can write it as $frac1-e^fracln(1-1/n)1/n+11/n$. Compute a few terms of the series expansion (in powers of $1/n$) of $fracln(1-1/n)1/n+1$ in powers of $1/n$ to get $-1/(2 n) - 1/(3 n^2)+...$. Then compose with the series of $e^x$ to get the initial terms of the series (in powers of $1/n$) of the original expression $1/2 + 5/(24 n)+...$. Therefore, the limit is $1/2$.
– spiralstotheleft
Jul 31 at 2:06
add a comment |Â
1 Answer
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beginalign
n-neleft(1-frac1nright)^n
&= n-nee^nlnleft(1-frac1nright) \
&= n-nee^left(-1-frac12n+O(frac1n^2)right) \
&= n-nleft(1-frac12n+O(frac1n^2)right) \
&= frac12+O(frac1n)
endalign
then
$$lim_ntoinfty n-neleft(1+frac1nright)^n=colorbluefrac12$$
answered Jul 31 at 2:07
user 108128
18.9k41544
Hah :) . . . . .
– user 108128
Jul 31 at 2:22
Only bright hearts are trusted.
– user 108128
Jul 31 at 2:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
beginalign
n-neleft(1-frac1nright)^n
&= n-nee^nlnleft(1-frac1nright) \
&= n-nee^left(-1-frac12n+O(frac1n^2)right) \
&= n-nleft(1-frac12n+O(frac1n^2)right) \
&= frac12+O(frac1n)
endalign
then
$$lim_ntoinfty n-neleft(1+frac1nright)^n=colorbluefrac12$$
answered Jul 31 at 2:07
user 108128
18.9k41544
Hah :) . . . . .
– user 108128
Jul 31 at 2:22
Only bright hearts are trusted.
– user 108128
Jul 31 at 2:51
add a comment |Â
up vote
1
down vote
accepted
beginalign
n-neleft(1-frac1nright)^n
&= n-nee^nlnleft(1-frac1nright) \
&= n-nee^left(-1-frac12n+O(frac1n^2)right) \
&= n-nleft(1-frac12n+O(frac1n^2)right) \
&= frac12+O(frac1n)
endalign
then
$$lim_ntoinfty n-neleft(1+frac1nright)^n=colorbluefrac12$$
answered Jul 31 at 2:07
user 108128
18.9k41544
Hah :) . . . . .
– user 108128
Jul 31 at 2:22
Only bright hearts are trusted.
– user 108128
Jul 31 at 2:51
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
beginalign
n-neleft(1-frac1nright)^n
&= n-nee^nlnleft(1-frac1nright) \
&= n-nee^left(-1-frac12n+O(frac1n^2)right) \
&= n-nleft(1-frac12n+O(frac1n^2)right) \
&= frac12+O(frac1n)
endalign
then
$$lim_ntoinfty n-neleft(1+frac1nright)^n=colorbluefrac12$$
answered Jul 31 at 2:07
user 108128
18.9k41544
beginalign
n-neleft(1-frac1nright)^n
&= n-nee^nlnleft(1-frac1nright) \
&= n-nee^left(-1-frac12n+O(frac1n^2)right) \
&= n-nleft(1-frac12n+O(frac1n^2)right) \
&= frac12+O(frac1n)
endalign
then
$$lim_ntoinfty n-neleft(1+frac1nright)^n=colorbluefrac12$$
answered Jul 31 at 2:07
user 108128
18.9k41544
edited Jul 31 at 2:12
answered Jul 31 at 2:07
user 108128
18.9k41544
answered Jul 31 at 2:07
user 108128
18.9k41544
answered Jul 31 at 2:07
user 108128
18.9k41544
18.9k41544
Hah :) . . . . .
– user 108128
Jul 31 at 2:22
Only bright hearts are trusted.
– user 108128
Jul 31 at 2:51
add a comment |Â
Hah :) . . . . .
– user 108128
Jul 31 at 2:22
Only bright hearts are trusted.
– user 108128
Jul 31 at 2:51
Hah :) . . . . .
– user 108128
Jul 31 at 2:22
Hah :) . . . . .
– user 108128
Jul 31 at 2:22
Only bright hearts are trusted.
– user 108128
Jul 31 at 2:51
Only bright hearts are trusted.
– user 108128
Jul 31 at 2:51
add a comment |Â
1
Write it as $n(1-e(1+1/n)^n)$, use that $lim_ntoinfty(1+1/n)^n=e$. Therefore $1-e(1+1/n)^nto1-e^2<0$. Finally, since $ntoinfty$, then the given limit tends to $-infty$.
– spiralstotheleft
Jul 31 at 1:46
1
For the new one you can write it as $frac1-e^fracln(1-1/n)1/n+11/n$. Compute a few terms of the series expansion (in powers of $1/n$) of $fracln(1-1/n)1/n+1$ in powers of $1/n$ to get $-1/(2 n) - 1/(3 n^2)+...$. Then compose with the series of $e^x$ to get the initial terms of the series (in powers of $1/n$) of the original expression $1/2 + 5/(24 n)+...$. Therefore, the limit is $1/2$.
– spiralstotheleft
Jul 31 at 2:06