Scary looking limit with an elegant answer Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraEvaluating 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
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraEvaluating 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 Apr 20 at 12:18
ZackyZacky
8,16511162
$endgroup$
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 Apr 20 at 12:18
ZackyZacky
8,16511162
$endgroup$
1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
Apr 20 at 12:23
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
Apr 20 at 12:34
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
Apr 20 at 12:40
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
Apr 20 at 12:59
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 Apr 20 at 12:18
ZackyZacky
8,16511162
$endgroup$
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 Apr 20 at 12:18
ZackyZacky
8,16511162
asked Apr 20 at 12:18
ZackyZacky
8,16511162
asked Apr 20 at 12:18
ZackyZacky
8,16511162
asked Apr 20 at 12:18
ZackyZacky
8,16511162
asked Apr 20 at 12:18
ZackyZacky
8,16511162
8,16511162
1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
Apr 20 at 12:23
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
Apr 20 at 12:34
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
Apr 20 at 12:40
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
Apr 20 at 12:59
add a comment |
1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
Apr 20 at 12:23
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
Apr 20 at 12:34
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
Apr 20 at 12:40
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
Apr 20 at 12:59
1
1
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
Apr 20 at 12:23
$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
$endgroup$
– Crostul
Apr 20 at 12:23
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
Apr 20 at 12:34
$begingroup$
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
Apr 20 at 12:34
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
Apr 20 at 12:40
$begingroup$
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
Apr 20 at 12:40
3
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
Apr 20 at 12:59
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
$endgroup$
– GEdgar
Apr 20 at 12:59
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 Apr 20 at 12:52
Paramanand SinghParamanand Singh
51.7k560171
$endgroup$
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
Apr 20 at 12:58
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
Apr 20 at 12:59
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
Apr 20 at 13:05
$begingroup$
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
Apr 20 at 13:14
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
Apr 20 at 13:15
add a comment |
Your Answer
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1 Answer
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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 Apr 20 at 12:52
Paramanand SinghParamanand Singh
51.7k560171
$endgroup$
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
Apr 20 at 12:58
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
Apr 20 at 12:59
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
Apr 20 at 13:05
$begingroup$
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
Apr 20 at 13:14
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
Apr 20 at 13:15
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 Apr 20 at 12:52
Paramanand SinghParamanand Singh
51.7k560171
$endgroup$
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
Apr 20 at 12:58
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
Apr 20 at 12:59
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
Apr 20 at 13:05
$begingroup$
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
Apr 20 at 13:14
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
Apr 20 at 13:15
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 Apr 20 at 12:52
Paramanand SinghParamanand Singh
51.7k560171
$endgroup$
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 Apr 20 at 12:52
Paramanand SinghParamanand Singh
51.7k560171
edited Apr 20 at 13:05
answered Apr 20 at 12:52
Paramanand SinghParamanand Singh
51.7k560171
answered Apr 20 at 12:52
Paramanand SinghParamanand Singh
51.7k560171
answered Apr 20 at 12:52
Paramanand SinghParamanand Singh
51.7k560171
51.7k560171
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
Apr 20 at 12:58
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
Apr 20 at 12:59
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
Apr 20 at 13:05
$begingroup$
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
Apr 20 at 13:14
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
Apr 20 at 13:15
add a comment |
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
Apr 20 at 12:58
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
Apr 20 at 12:59
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
$endgroup$
– Paramanand Singh
Apr 20 at 13:05
$begingroup$
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
Apr 20 at 13:14
1
$begingroup$
@Zacky: I am sure it would appear damn scary to many people.
$endgroup$
– Paramanand Singh
Apr 20 at 13:15
1
1
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
Apr 20 at 12:58
$begingroup$
And "$(n!) ^2k/(2kn)!$ tends to zero" does not require Stirling's formula.
$endgroup$
– GEdgar
Apr 20 at 12:58
1
1
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@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
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– Paramanand Singh
Apr 20 at 12:59
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@GEdgar: I did not check that but it appears one can use ratio test or something similar to conclude that it tends to $0$.
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– Paramanand Singh
Apr 20 at 12:59
1
1
$begingroup$
@GEdgar: finally added a demonstration via ratio test.
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– Paramanand Singh
Apr 20 at 13:05
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@GEdgar: finally added a demonstration via ratio test.
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– Paramanand Singh
Apr 20 at 13:05
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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
Apr 20 at 13:14
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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
Apr 20 at 13:14
1
1
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@Zacky: I am sure it would appear damn scary to many people.
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– Paramanand Singh
Apr 20 at 13:15
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@Zacky: I am sure it would appear damn scary to many people.
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– Paramanand Singh
Apr 20 at 13:15
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$begingroup$
Maybe it could be a good idea to use Stirling, just to spot what the limit is?
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– Crostul
Apr 20 at 12:23
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.
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– Paramanand Singh
Apr 20 at 12:34
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
Apr 20 at 12:40
3
$begingroup$
A pedantic answerer could go ahead and use Stirling's formula, since the forbidden usage is misspelled Strirling's.
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– GEdgar
Apr 20 at 12:59