Scary looking limit with an elegant answer Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Evaluating a limit with variable in the exponentFinding a limit with squeeze theoremCalculate the limit.Limit of function with 0/0Evaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Limit using l'hospitalProblem with finding a limit of a tricky functionCan I take limits under a limit?Limit answer not matchingLimit $lim_ntoinfty n^2left(sqrt1+frac1n+sqrt1-frac1n-2right)$
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Scary looking limit with an elegant answer
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Evaluating a limit with variable in the exponentFinding a limit with squeeze theoremCalculate the limit.Limit of function with 0/0Evaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Limit using l'hospitalProblem with finding a limit of a tricky functionCan I take limits under a limit?Limit answer not matchingLimit $lim_ntoinfty n^2left(sqrt1+frac1n+sqrt1-frac1n-2right)$
$begingroup$
This problem was posted half a year ago by Pierre Mounir on a Facebook group and until now it received no answers. Since most of his problems that I saw were amazing I can bet this one it's worth the time. Wolfram returns the answer to be $2$, which is quite elegant for it's look.
I remembered about it yesterday and gave a try again taking for simplicity $k=1$ (I had no chance with a bigger number).
Also my whole idea was to somehow get to a point where I can use $limlimits_fto 0fraca^f-1f=ln a$, thus I started as:
$$lim_nto infty frac5^fracn!(2n)!-4^frac1n!3^fracn!(2n)!-2^frac1n!=lim_nto infty left(frac42right)^frac1n!left(frac5^fracn!(2n)!4^frac1n!-1right)left(frac3^fracn!(2n)!2^frac1n!-1right)^-1$$
$$=lim_nto infty underbracesqrt[n!]2_to 1left(sqrt[n!]frac5^frac1(2n)!4-1right)left(sqrt[n!]frac3^frac1(2n)!2-1right)^-1$$
Well, yes $frac5^frac1(2n)!4$ and the other one in the denominator equals to $1$, but still I don't see how to use that limit. Also I tried to take a logarithm on both sides or to use L'hospital, but looks like a dead end.
I would love if someone can spot the trick for solving this limit and land some help.
limits factorial
asked yesterday
ZackyZacky
7,98711062
$endgroup$
add a comment |
$begingroup$
This problem was posted half a year ago by Pierre Mounir on a Facebook group and until now it received no answers. Since most of his problems that I saw were amazing I can bet this one it's worth the time. Wolfram returns the answer to be $2$, which is quite elegant for it's look.
I remembered about it yesterday and gave a try again taking for simplicity $k=1$ (I had no chance with a bigger number).
Also my whole idea was to somehow get to a point where I can use $limlimits_fto 0fraca^f-1f=ln a$, thus I started as:
$$lim_nto infty frac5^fracn!(2n)!-4^frac1n!3^fracn!(2n)!-2^frac1n!=lim_nto infty left(frac42right)^frac1n!left(frac5^fracn!(2n)!4^frac1n!-1right)left(frac3^fracn!(2n)!2^frac1n!-1right)^-1$$
$$=lim_nto infty underbracesqrt[n!]2_to 1left(sqrt[n!]frac5^frac1(2n)!4-1right)left(sqrt[n!]frac3^frac1(2n)!2-1right)^-1$$
Well, yes $frac5^frac1(2n)!4$ and the other one in the denominator equals to $1$, but still I don't see how to use that limit. Also I tried to take a logarithm on both sides or to use L'hospital, but looks like a dead end.
I would love if someone can spot the trick for solving this limit and land some help.
limits factorial
asked yesterday
ZackyZacky
7,98711062
$endgroup$
1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
yesterday
2
$begingroup$
If the given expression is $(a-b) /(c-d) $ then $a, b, c, d$ tend to $1$ and hence they can be replaced by their logs.
$endgroup$
– Paramanand Singh
yesterday
3
$begingroup$
The answer easily comes to be 2 via logs. One just needs to prove that $(n!) ^2 k /(2kn)!to 0$ which is sort of not that difficult.
$endgroup$
– Paramanand Singh
yesterday
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
yesterday
add a comment |
$begingroup$
This problem was posted half a year ago by Pierre Mounir on a Facebook group and until now it received no answers. Since most of his problems that I saw were amazing I can bet this one it's worth the time. Wolfram returns the answer to be $2$, which is quite elegant for it's look.
I remembered about it yesterday and gave a try again taking for simplicity $k=1$ (I had no chance with a bigger number).
Also my whole idea was to somehow get to a point where I can use $limlimits_fto 0fraca^f-1f=ln a$, thus I started as:
$$lim_nto infty frac5^fracn!(2n)!-4^frac1n!3^fracn!(2n)!-2^frac1n!=lim_nto infty left(frac42right)^frac1n!left(frac5^fracn!(2n)!4^frac1n!-1right)left(frac3^fracn!(2n)!2^frac1n!-1right)^-1$$
$$=lim_nto infty underbracesqrt[n!]2_to 1left(sqrt[n!]frac5^frac1(2n)!4-1right)left(sqrt[n!]frac3^frac1(2n)!2-1right)^-1$$
Well, yes $frac5^frac1(2n)!4$ and the other one in the denominator equals to $1$, but still I don't see how to use that limit. Also I tried to take a logarithm on both sides or to use L'hospital, but looks like a dead end.
I would love if someone can spot the trick for solving this limit and land some help.
limits factorial
asked yesterday
ZackyZacky
7,98711062
$endgroup$
This problem was posted half a year ago by Pierre Mounir on a Facebook group and until now it received no answers. Since most of his problems that I saw were amazing I can bet this one it's worth the time. Wolfram returns the answer to be $2$, which is quite elegant for it's look.
I remembered about it yesterday and gave a try again taking for simplicity $k=1$ (I had no chance with a bigger number).
Also my whole idea was to somehow get to a point where I can use $limlimits_fto 0fraca^f-1f=ln a$, thus I started as:
$$lim_nto infty frac5^fracn!(2n)!-4^frac1n!3^fracn!(2n)!-2^frac1n!=lim_nto infty left(frac42right)^frac1n!left(frac5^fracn!(2n)!4^frac1n!-1right)left(frac3^fracn!(2n)!2^frac1n!-1right)^-1$$
$$=lim_nto infty underbracesqrt[n!]2_to 1left(sqrt[n!]frac5^frac1(2n)!4-1right)left(sqrt[n!]frac3^frac1(2n)!2-1right)^-1$$
Well, yes $frac5^frac1(2n)!4$ and the other one in the denominator equals to $1$, but still I don't see how to use that limit. Also I tried to take a logarithm on both sides or to use L'hospital, but looks like a dead end.
I would love if someone can spot the trick for solving this limit and land some help.
limits factorial
limits factorial
asked yesterday
ZackyZacky
7,98711062
asked yesterday
ZackyZacky
7,98711062
asked yesterday
ZackyZacky
7,98711062
asked yesterday
ZackyZacky
7,98711062
asked yesterday
ZackyZacky
7,98711062
7,98711062
1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
yesterday
2
$begingroup$
If the given expression is $(a-b) /(c-d) $ then $a, b, c, d$ tend to $1$ and hence they can be replaced by their logs.
$endgroup$
– Paramanand Singh
yesterday
3
$begingroup$
The answer easily comes to be 2 via logs. One just needs to prove that $(n!) ^2 k /(2kn)!to 0$ which is sort of not that difficult.
$endgroup$
– Paramanand Singh
yesterday
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
yesterday
add a comment |
1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
yesterday
2
$begingroup$
If the given expression is $(a-b) /(c-d) $ then $a, b, c, d$ tend to $1$ and hence they can be replaced by their logs.
$endgroup$
– Paramanand Singh
yesterday
3
$begingroup$
The answer easily comes to be 2 via logs. One just needs to prove that $(n!) ^2 k /(2kn)!to 0$ which is sort of not that difficult.
$endgroup$
– Paramanand Singh
yesterday
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
yesterday
1
1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
yesterday
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
yesterday
2
2
$begingroup$
If the given expression is $(a-b) /(c-d) $ then $a, b, c, d$ tend to $1$ and hence they can be replaced by their logs.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
If the given expression is $(a-b) /(c-d) $ then $a, b, c, d$ tend to $1$ and hence they can be replaced by their logs.
$endgroup$
– Paramanand Singh
yesterday
3
3
$begingroup$
The answer easily comes to be 2 via logs. One just needs to prove that $(n!) ^2 k /(2kn)!to 0$ which is sort of not that difficult.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
The answer easily comes to be 2 via logs. One just needs to prove that $(n!) ^2 k /(2kn)!to 0$ which is sort of not that difficult.
$endgroup$
– Paramanand Singh
yesterday
3
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
yesterday
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is an expansion of my comment into an answer.
If the expression under limit is of the form $(A-B) /(C-D) $ then all of $A, B, C, D$ tend to $1$ and we can write $$A-B=Bcdotfracexp(log A-log B) - 1log A - log B cdot(log A - log B) $$ The first two factors tend to $1$ and hence the numerator can be replaced by $log A-log B$. Proceeding in similar manner we see that the expression under limit can be simplified greatly as each term gets replaced by its logarithm.
The resulting expression is $$fracalog 5-2blog 2alog 3 - blog 2$$ where $a=(n!) ^k /(2kn)!,b=(n!)^-k$ and clearly $a/b=(n!) ^2k/(2kn)!$ tends to $0$ so that the desired limit is $2$.
The limiting behavior of $a_n=(n!) ^2 k /(2kn)!$ can be concluded via ratio test. We have $$fraca_n+1a_n =frac(n+1)^2k(2kn+1)cdots(2kn+2k)tofrac1(2k)^2k<1$$ and hence $a_nto 0$.
answered yesterday
Paramanand SinghParamanand Singh
51.6k560170
$endgroup$
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
yesterday
1
$begingroup$
@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
$endgroup$
– Paramanand Singh
yesterday
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
Uff, I should remove scary from the name since you made it look easy. I didn't realise how useful is to write $frac5^fracn!(2n)!4^frac1n!$ in the $e^log $ form. There I too tried to get into that special limit from, but taking into $limlimits_f(x)to 0fraca^f(x)-1f(x),,$ the $f(x)$ term as $log A-log B$ was clearly what I needed, thanks alot!
$endgroup$
– Zacky
yesterday
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
yesterday
add a comment |
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$begingroup$
This is an expansion of my comment into an answer.
If the expression under limit is of the form $(A-B) /(C-D) $ then all of $A, B, C, D$ tend to $1$ and we can write $$A-B=Bcdotfracexp(log A-log B) - 1log A - log B cdot(log A - log B) $$ The first two factors tend to $1$ and hence the numerator can be replaced by $log A-log B$. Proceeding in similar manner we see that the expression under limit can be simplified greatly as each term gets replaced by its logarithm.
The resulting expression is $$fracalog 5-2blog 2alog 3 - blog 2$$ where $a=(n!) ^k /(2kn)!,b=(n!)^-k$ and clearly $a/b=(n!) ^2k/(2kn)!$ tends to $0$ so that the desired limit is $2$.
The limiting behavior of $a_n=(n!) ^2 k /(2kn)!$ can be concluded via ratio test. We have $$fraca_n+1a_n =frac(n+1)^2k(2kn+1)cdots(2kn+2k)tofrac1(2k)^2k<1$$ and hence $a_nto 0$.
answered yesterday
Paramanand SinghParamanand Singh
51.6k560170
$endgroup$
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
yesterday
1
$begingroup$
@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
$endgroup$
– Paramanand Singh
yesterday
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
Uff, I should remove scary from the name since you made it look easy. I didn't realise how useful is to write $frac5^fracn!(2n)!4^frac1n!$ in the $e^log $ form. There I too tried to get into that special limit from, but taking into $limlimits_f(x)to 0fraca^f(x)-1f(x),,$ the $f(x)$ term as $log A-log B$ was clearly what I needed, thanks alot!
$endgroup$
– Zacky
yesterday
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
yesterday
add a comment |
$begingroup$
This is an expansion of my comment into an answer.
If the expression under limit is of the form $(A-B) /(C-D) $ then all of $A, B, C, D$ tend to $1$ and we can write $$A-B=Bcdotfracexp(log A-log B) - 1log A - log B cdot(log A - log B) $$ The first two factors tend to $1$ and hence the numerator can be replaced by $log A-log B$. Proceeding in similar manner we see that the expression under limit can be simplified greatly as each term gets replaced by its logarithm.
The resulting expression is $$fracalog 5-2blog 2alog 3 - blog 2$$ where $a=(n!) ^k /(2kn)!,b=(n!)^-k$ and clearly $a/b=(n!) ^2k/(2kn)!$ tends to $0$ so that the desired limit is $2$.
The limiting behavior of $a_n=(n!) ^2 k /(2kn)!$ can be concluded via ratio test. We have $$fraca_n+1a_n =frac(n+1)^2k(2kn+1)cdots(2kn+2k)tofrac1(2k)^2k<1$$ and hence $a_nto 0$.
answered yesterday
Paramanand SinghParamanand Singh
51.6k560170
$endgroup$
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
yesterday
1
$begingroup$
@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
$endgroup$
– Paramanand Singh
yesterday
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
Uff, I should remove scary from the name since you made it look easy. I didn't realise how useful is to write $frac5^fracn!(2n)!4^frac1n!$ in the $e^log $ form. There I too tried to get into that special limit from, but taking into $limlimits_f(x)to 0fraca^f(x)-1f(x),,$ the $f(x)$ term as $log A-log B$ was clearly what I needed, thanks alot!
$endgroup$
– Zacky
yesterday
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
yesterday
add a comment |
$begingroup$
This is an expansion of my comment into an answer.
If the expression under limit is of the form $(A-B) /(C-D) $ then all of $A, B, C, D$ tend to $1$ and we can write $$A-B=Bcdotfracexp(log A-log B) - 1log A - log B cdot(log A - log B) $$ The first two factors tend to $1$ and hence the numerator can be replaced by $log A-log B$. Proceeding in similar manner we see that the expression under limit can be simplified greatly as each term gets replaced by its logarithm.
The resulting expression is $$fracalog 5-2blog 2alog 3 - blog 2$$ where $a=(n!) ^k /(2kn)!,b=(n!)^-k$ and clearly $a/b=(n!) ^2k/(2kn)!$ tends to $0$ so that the desired limit is $2$.
The limiting behavior of $a_n=(n!) ^2 k /(2kn)!$ can be concluded via ratio test. We have $$fraca_n+1a_n =frac(n+1)^2k(2kn+1)cdots(2kn+2k)tofrac1(2k)^2k<1$$ and hence $a_nto 0$.
answered yesterday
Paramanand SinghParamanand Singh
51.6k560170
$endgroup$
This is an expansion of my comment into an answer.
If the expression under limit is of the form $(A-B) /(C-D) $ then all of $A, B, C, D$ tend to $1$ and we can write $$A-B=Bcdotfracexp(log A-log B) - 1log A - log B cdot(log A - log B) $$ The first two factors tend to $1$ and hence the numerator can be replaced by $log A-log B$. Proceeding in similar manner we see that the expression under limit can be simplified greatly as each term gets replaced by its logarithm.
The resulting expression is $$fracalog 5-2blog 2alog 3 - blog 2$$ where $a=(n!) ^k /(2kn)!,b=(n!)^-k$ and clearly $a/b=(n!) ^2k/(2kn)!$ tends to $0$ so that the desired limit is $2$.
The limiting behavior of $a_n=(n!) ^2 k /(2kn)!$ can be concluded via ratio test. We have $$fraca_n+1a_n =frac(n+1)^2k(2kn+1)cdots(2kn+2k)tofrac1(2k)^2k<1$$ and hence $a_nto 0$.
answered yesterday
Paramanand SinghParamanand Singh
51.6k560170
edited yesterday
answered yesterday
Paramanand SinghParamanand Singh
51.6k560170
answered yesterday
Paramanand SinghParamanand Singh
51.6k560170
answered yesterday
Paramanand SinghParamanand Singh
51.6k560170
51.6k560170
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
yesterday
1
$begingroup$
@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
$endgroup$
– Paramanand Singh
yesterday
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
Uff, I should remove scary from the name since you made it look easy. I didn't realise how useful is to write $frac5^fracn!(2n)!4^frac1n!$ in the $e^log $ form. There I too tried to get into that special limit from, but taking into $limlimits_f(x)to 0fraca^f(x)-1f(x),,$ the $f(x)$ term as $log A-log B$ was clearly what I needed, thanks alot!
$endgroup$
– Zacky
yesterday
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
yesterday
add a comment |
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
yesterday
1
$begingroup$
@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
$endgroup$
– Paramanand Singh
yesterday
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
Uff, I should remove scary from the name since you made it look easy. I didn't realise how useful is to write $frac5^fracn!(2n)!4^frac1n!$ in the $e^log $ form. There I too tried to get into that special limit from, but taking into $limlimits_f(x)to 0fraca^f(x)-1f(x),,$ the $f(x)$ term as $log A-log B$ was clearly what I needed, thanks alot!
$endgroup$
– Zacky
yesterday
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
yesterday
1
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
yesterday
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
yesterday
1
1
$begingroup$
@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
$endgroup$
– Paramanand Singh
yesterday
1
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
yesterday
$begingroup$
Uff, I should remove scary from the name since you made it look easy. I didn't realise how useful is to write $frac5^fracn!(2n)!4^frac1n!$ in the $e^log $ form. There I too tried to get into that special limit from, but taking into $limlimits_f(x)to 0fraca^f(x)-1f(x),,$ the $f(x)$ term as $log A-log B$ was clearly what I needed, thanks alot!
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– Zacky
yesterday
$begingroup$
Uff, I should remove scary from the name since you made it look easy. I didn't realise how useful is to write $frac5^fracn!(2n)!4^frac1n!$ in the $e^log $ form. There I too tried to get into that special limit from, but taking into $limlimits_f(x)to 0fraca^f(x)-1f(x),,$ the $f(x)$ term as $log A-log B$ was clearly what I needed, thanks alot!
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– Zacky
yesterday
1
1
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@Zacky: I am sure it would appear damn scary to many people.
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– Paramanand Singh
yesterday
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@Zacky: I am sure it would appear damn scary to many people.
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– Paramanand Singh
yesterday
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1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
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– Crostul
yesterday
2
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If the given expression is $(a-b) /(c-d) $ then $a, b, c, d$ tend to $1$ and hence they can be replaced by their logs.
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– Paramanand Singh
yesterday
3
$begingroup$
The answer easily comes to be 2 via logs. One just needs to prove that $(n!) ^2 k /(2kn)!to 0$ which is sort of not that difficult.
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– Paramanand Singh
yesterday
3
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A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
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– GEdgar
yesterday