Find $lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{{(x)}^3}}{x!^{(1/x)}}right)$












2












$begingroup$


$$lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}right)$$



I have been trying to find the limit of the following sequence above but I am stuck. I first tried using substitution for $sin(1/x)$ because it approaches $0$; however, that would make the entire expression $0$ which is obviously wrong.



I next tried using the laws of logs, but that just made it more complicated getting nowhere.






calculus limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
    $endgroup$
    – Andrew Shedlock
    Jan 3 at 1:26










  • $begingroup$
    The denominator I think is asymptotic to $x/e$ (for example using Stirling's asymptotic formula), but numerator seems to be asymtptotic to $1/6$, which would make the limit $0$. If I replace $x^3$ with $x^4$ at the numerator then, I get $e/6$ as limit... Does it make any sense to you? If it does I can show you details of my calculations.
    $endgroup$
    – Matteo
    Jan 3 at 1:34












  • $begingroup$
    We haven't learned the Gamma function, yet. Thank you though.
    $endgroup$
    – David MB
    Jan 3 at 2:56
















2












$begingroup$


$$lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}right)$$



I have been trying to find the limit of the following sequence above but I am stuck. I first tried using substitution for $sin(1/x)$ because it approaches $0$; however, that would make the entire expression $0$ which is obviously wrong.



I next tried using the laws of logs, but that just made it more complicated getting nowhere.






calculus limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
    $endgroup$
    – Andrew Shedlock
    Jan 3 at 1:26










  • $begingroup$
    The denominator I think is asymptotic to $x/e$ (for example using Stirling's asymptotic formula), but numerator seems to be asymtptotic to $1/6$, which would make the limit $0$. If I replace $x^3$ with $x^4$ at the numerator then, I get $e/6$ as limit... Does it make any sense to you? If it does I can show you details of my calculations.
    $endgroup$
    – Matteo
    Jan 3 at 1:34












  • $begingroup$
    We haven't learned the Gamma function, yet. Thank you though.
    $endgroup$
    – David MB
    Jan 3 at 2:56














2












2








2


2



$begingroup$


$$lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}right)$$



I have been trying to find the limit of the following sequence above but I am stuck. I first tried using substitution for $sin(1/x)$ because it approaches $0$; however, that would make the entire expression $0$ which is obviously wrong.



I next tried using the laws of logs, but that just made it more complicated getting nowhere.






calculus limits







share|cite|improve this question











$endgroup$




$$lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}right)$$



I have been trying to find the limit of the following sequence above but I am stuck. I first tried using substitution for $sin(1/x)$ because it approaches $0$; however, that would make the entire expression $0$ which is obviously wrong.



I next tried using the laws of logs, but that just made it more complicated getting nowhere.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 9:24









zhw.

72.4k43175




72.4k43175










asked Jan 3 at 0:25









David MBDavid MB

287




287








  • 1




    $begingroup$
    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
    $endgroup$
    – Andrew Shedlock
    Jan 3 at 1:26










  • $begingroup$
    The denominator I think is asymptotic to $x/e$ (for example using Stirling's asymptotic formula), but numerator seems to be asymtptotic to $1/6$, which would make the limit $0$. If I replace $x^3$ with $x^4$ at the numerator then, I get $e/6$ as limit... Does it make any sense to you? If it does I can show you details of my calculations.
    $endgroup$
    – Matteo
    Jan 3 at 1:34












  • $begingroup$
    We haven't learned the Gamma function, yet. Thank you though.
    $endgroup$
    – David MB
    Jan 3 at 2:56














  • 1




    $begingroup$
    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
    $endgroup$
    – Andrew Shedlock
    Jan 3 at 1:26










  • $begingroup$
    The denominator I think is asymptotic to $x/e$ (for example using Stirling's asymptotic formula), but numerator seems to be asymtptotic to $1/6$, which would make the limit $0$. If I replace $x^3$ with $x^4$ at the numerator then, I get $e/6$ as limit... Does it make any sense to you? If it does I can show you details of my calculations.
    $endgroup$
    – Matteo
    Jan 3 at 1:34












  • $begingroup$
    We haven't learned the Gamma function, yet. Thank you though.
    $endgroup$
    – David MB
    Jan 3 at 2:56








1




1




$begingroup$
Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
$endgroup$
– Andrew Shedlock
Jan 3 at 1:26




$begingroup$
Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
$endgroup$
– Andrew Shedlock
Jan 3 at 1:26












$begingroup$
The denominator I think is asymptotic to $x/e$ (for example using Stirling's asymptotic formula), but numerator seems to be asymtptotic to $1/6$, which would make the limit $0$. If I replace $x^3$ with $x^4$ at the numerator then, I get $e/6$ as limit... Does it make any sense to you? If it does I can show you details of my calculations.
$endgroup$
– Matteo
Jan 3 at 1:34






$begingroup$
The denominator I think is asymptotic to $x/e$ (for example using Stirling's asymptotic formula), but numerator seems to be asymtptotic to $1/6$, which would make the limit $0$. If I replace $x^3$ with $x^4$ at the numerator then, I get $e/6$ as limit... Does it make any sense to you? If it does I can show you details of my calculations.
$endgroup$
– Matteo
Jan 3 at 1:34














$begingroup$
We haven't learned the Gamma function, yet. Thank you though.
$endgroup$
– David MB
Jan 3 at 2:56




$begingroup$
We haven't learned the Gamma function, yet. Thank you though.
$endgroup$
– David MB
Jan 3 at 2:56










2 Answers
2






active

oldest

votes


















1












$begingroup$

begin{align}
&lim_{x to infty}frac{(exp(1/x) - exp(sin(1/x)))x^3}{(x!)^frac1x} \&= lim_{x to infty}frac{(1+frac1x+ frac1{2x^2} +frac1{6x^3}- exp(frac1x-frac1{6x^3}))x^3}{(x!)^frac1x}\
&=lim_{x to infty}frac{(1+frac1x+ frac1{2x^2}+frac1{6x^3} - left(1+frac1x+frac1{2x^2} -frac1{6x^3}+frac1{6x^3}right))x^3)}{(x!)^frac1x}\
&=frac16 cdot lim_{x to infty} frac1{expleft( frac{ln x!}{x}right)}\
&= 0
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    $endgroup$
    – David MB
    Jan 3 at 2:50












  • $begingroup$
    I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    $endgroup$
    – Siong Thye Goh
    Jan 3 at 2:58










  • $begingroup$
    I see. I'll definitely keep that in mind next time.
    $endgroup$
    – David MB
    Jan 3 at 3:17



















3












$begingroup$

First note that $x/(x!) ^{1/x}to e$ where by $x! $ we mean $Gamma (x+1)$ and hence the desired limit is equal to the limit of the expression $$ecdot(e^{1/x}-e^{sin(1/x)})x^2$$ Putting $x=1/t$ so that $tto 0^{+}$ we see that the desired limit is $$elim_{tto 0^{+}}e^{sin t} cdotfrac{e^{t-sin t} - 1}{t-sin t} cdot frac{t-sin t} {t^2}$$ Clearly the first two factors under limit operation tend to $1$ and the last one tends to $0$ and thus the limit in question is $0$.




The fact that $x/Gamma ^{1/x}(x+1)$ tends to $e$ as $xtoinfty $ can be established in the following manner.



Let $f(x) =log(x^x/Gamma (x+1))$ then $$f(x+1)-f(x)=xlogleft(1+frac {1}{x}right) $$ which tends to $1$ and hence $f(x) /xto 1$. And thus $log (x/Gamma ^{1/x}(x+1))to 1$.




The case when $x$ is an integer is famous and we can prove that $limlimits _{ntoinfty} dfrac{n} {sqrt[n] {n!}} =e$ using the well known




Theorem: If ${a_n} $ is a sequence of positive terms such that $a_{n+1}/a_nto L$ as $ntoinfty $ then $sqrt[n] {a_n} to L$ as $ntoinfty $.




Taking $a_n=n^n/n! $ we can see that $$frac{a_{n+1}}{a_n}=left(1+frac {1}{n}right) ^nto e$$ so that $sqrt[n] {a_n} =n/sqrt[n] {n!} to e$.



A similar theorem like the one above holds for functions of a real variable and the same has been used in my solution which uses $x$ as a real variable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your solution. I am curious of how did you know it tends to e and what gave you the inclination to use substitution?
    $endgroup$
    – David MB
    Jan 3 at 6:58












  • $begingroup$
    @DavidMB: the limit $lim_{ntoinfty} n/sqrt[n] {n!} =e$ ($n$ integer) is a famous exercise and is already available on this website. I have given a similar argument replacing integer $n$ with real variable $x$.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:07










  • $begingroup$
    @DavidMB: I have added an update in my answer with some details.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:22











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

begin{align}
&lim_{x to infty}frac{(exp(1/x) - exp(sin(1/x)))x^3}{(x!)^frac1x} \&= lim_{x to infty}frac{(1+frac1x+ frac1{2x^2} +frac1{6x^3}- exp(frac1x-frac1{6x^3}))x^3}{(x!)^frac1x}\
&=lim_{x to infty}frac{(1+frac1x+ frac1{2x^2}+frac1{6x^3} - left(1+frac1x+frac1{2x^2} -frac1{6x^3}+frac1{6x^3}right))x^3)}{(x!)^frac1x}\
&=frac16 cdot lim_{x to infty} frac1{expleft( frac{ln x!}{x}right)}\
&= 0
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    $endgroup$
    – David MB
    Jan 3 at 2:50












  • $begingroup$
    I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    $endgroup$
    – Siong Thye Goh
    Jan 3 at 2:58










  • $begingroup$
    I see. I'll definitely keep that in mind next time.
    $endgroup$
    – David MB
    Jan 3 at 3:17
















1












$begingroup$

begin{align}
&lim_{x to infty}frac{(exp(1/x) - exp(sin(1/x)))x^3}{(x!)^frac1x} \&= lim_{x to infty}frac{(1+frac1x+ frac1{2x^2} +frac1{6x^3}- exp(frac1x-frac1{6x^3}))x^3}{(x!)^frac1x}\
&=lim_{x to infty}frac{(1+frac1x+ frac1{2x^2}+frac1{6x^3} - left(1+frac1x+frac1{2x^2} -frac1{6x^3}+frac1{6x^3}right))x^3)}{(x!)^frac1x}\
&=frac16 cdot lim_{x to infty} frac1{expleft( frac{ln x!}{x}right)}\
&= 0
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    $endgroup$
    – David MB
    Jan 3 at 2:50












  • $begingroup$
    I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    $endgroup$
    – Siong Thye Goh
    Jan 3 at 2:58










  • $begingroup$
    I see. I'll definitely keep that in mind next time.
    $endgroup$
    – David MB
    Jan 3 at 3:17














1












1








1





$begingroup$

begin{align}
&lim_{x to infty}frac{(exp(1/x) - exp(sin(1/x)))x^3}{(x!)^frac1x} \&= lim_{x to infty}frac{(1+frac1x+ frac1{2x^2} +frac1{6x^3}- exp(frac1x-frac1{6x^3}))x^3}{(x!)^frac1x}\
&=lim_{x to infty}frac{(1+frac1x+ frac1{2x^2}+frac1{6x^3} - left(1+frac1x+frac1{2x^2} -frac1{6x^3}+frac1{6x^3}right))x^3)}{(x!)^frac1x}\
&=frac16 cdot lim_{x to infty} frac1{expleft( frac{ln x!}{x}right)}\
&= 0
end{align}






share|cite|improve this answer











$endgroup$



begin{align}
&lim_{x to infty}frac{(exp(1/x) - exp(sin(1/x)))x^3}{(x!)^frac1x} \&= lim_{x to infty}frac{(1+frac1x+ frac1{2x^2} +frac1{6x^3}- exp(frac1x-frac1{6x^3}))x^3}{(x!)^frac1x}\
&=lim_{x to infty}frac{(1+frac1x+ frac1{2x^2}+frac1{6x^3} - left(1+frac1x+frac1{2x^2} -frac1{6x^3}+frac1{6x^3}right))x^3)}{(x!)^frac1x}\
&=frac16 cdot lim_{x to infty} frac1{expleft( frac{ln x!}{x}right)}\
&= 0
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 2:21

























answered Jan 3 at 1:56









Siong Thye GohSiong Thye Goh

101k1466117




101k1466117












  • $begingroup$
    Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    $endgroup$
    – David MB
    Jan 3 at 2:50












  • $begingroup$
    I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    $endgroup$
    – Siong Thye Goh
    Jan 3 at 2:58










  • $begingroup$
    I see. I'll definitely keep that in mind next time.
    $endgroup$
    – David MB
    Jan 3 at 3:17


















  • $begingroup$
    Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    $endgroup$
    – David MB
    Jan 3 at 2:50












  • $begingroup$
    I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    $endgroup$
    – Siong Thye Goh
    Jan 3 at 2:58










  • $begingroup$
    I see. I'll definitely keep that in mind next time.
    $endgroup$
    – David MB
    Jan 3 at 3:17
















$begingroup$
Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
$endgroup$
– David MB
Jan 3 at 2:50






$begingroup$
Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
$endgroup$
– David MB
Jan 3 at 2:50














$begingroup$
I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
$endgroup$
– Siong Thye Goh
Jan 3 at 2:58




$begingroup$
I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
$endgroup$
– Siong Thye Goh
Jan 3 at 2:58












$begingroup$
I see. I'll definitely keep that in mind next time.
$endgroup$
– David MB
Jan 3 at 3:17




$begingroup$
I see. I'll definitely keep that in mind next time.
$endgroup$
– David MB
Jan 3 at 3:17











3












$begingroup$

First note that $x/(x!) ^{1/x}to e$ where by $x! $ we mean $Gamma (x+1)$ and hence the desired limit is equal to the limit of the expression $$ecdot(e^{1/x}-e^{sin(1/x)})x^2$$ Putting $x=1/t$ so that $tto 0^{+}$ we see that the desired limit is $$elim_{tto 0^{+}}e^{sin t} cdotfrac{e^{t-sin t} - 1}{t-sin t} cdot frac{t-sin t} {t^2}$$ Clearly the first two factors under limit operation tend to $1$ and the last one tends to $0$ and thus the limit in question is $0$.




The fact that $x/Gamma ^{1/x}(x+1)$ tends to $e$ as $xtoinfty $ can be established in the following manner.



Let $f(x) =log(x^x/Gamma (x+1))$ then $$f(x+1)-f(x)=xlogleft(1+frac {1}{x}right) $$ which tends to $1$ and hence $f(x) /xto 1$. And thus $log (x/Gamma ^{1/x}(x+1))to 1$.




The case when $x$ is an integer is famous and we can prove that $limlimits _{ntoinfty} dfrac{n} {sqrt[n] {n!}} =e$ using the well known




Theorem: If ${a_n} $ is a sequence of positive terms such that $a_{n+1}/a_nto L$ as $ntoinfty $ then $sqrt[n] {a_n} to L$ as $ntoinfty $.




Taking $a_n=n^n/n! $ we can see that $$frac{a_{n+1}}{a_n}=left(1+frac {1}{n}right) ^nto e$$ so that $sqrt[n] {a_n} =n/sqrt[n] {n!} to e$.



A similar theorem like the one above holds for functions of a real variable and the same has been used in my solution which uses $x$ as a real variable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your solution. I am curious of how did you know it tends to e and what gave you the inclination to use substitution?
    $endgroup$
    – David MB
    Jan 3 at 6:58












  • $begingroup$
    @DavidMB: the limit $lim_{ntoinfty} n/sqrt[n] {n!} =e$ ($n$ integer) is a famous exercise and is already available on this website. I have given a similar argument replacing integer $n$ with real variable $x$.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:07










  • $begingroup$
    @DavidMB: I have added an update in my answer with some details.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:22
















3












$begingroup$

First note that $x/(x!) ^{1/x}to e$ where by $x! $ we mean $Gamma (x+1)$ and hence the desired limit is equal to the limit of the expression $$ecdot(e^{1/x}-e^{sin(1/x)})x^2$$ Putting $x=1/t$ so that $tto 0^{+}$ we see that the desired limit is $$elim_{tto 0^{+}}e^{sin t} cdotfrac{e^{t-sin t} - 1}{t-sin t} cdot frac{t-sin t} {t^2}$$ Clearly the first two factors under limit operation tend to $1$ and the last one tends to $0$ and thus the limit in question is $0$.




The fact that $x/Gamma ^{1/x}(x+1)$ tends to $e$ as $xtoinfty $ can be established in the following manner.



Let $f(x) =log(x^x/Gamma (x+1))$ then $$f(x+1)-f(x)=xlogleft(1+frac {1}{x}right) $$ which tends to $1$ and hence $f(x) /xto 1$. And thus $log (x/Gamma ^{1/x}(x+1))to 1$.




The case when $x$ is an integer is famous and we can prove that $limlimits _{ntoinfty} dfrac{n} {sqrt[n] {n!}} =e$ using the well known




Theorem: If ${a_n} $ is a sequence of positive terms such that $a_{n+1}/a_nto L$ as $ntoinfty $ then $sqrt[n] {a_n} to L$ as $ntoinfty $.




Taking $a_n=n^n/n! $ we can see that $$frac{a_{n+1}}{a_n}=left(1+frac {1}{n}right) ^nto e$$ so that $sqrt[n] {a_n} =n/sqrt[n] {n!} to e$.



A similar theorem like the one above holds for functions of a real variable and the same has been used in my solution which uses $x$ as a real variable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your solution. I am curious of how did you know it tends to e and what gave you the inclination to use substitution?
    $endgroup$
    – David MB
    Jan 3 at 6:58












  • $begingroup$
    @DavidMB: the limit $lim_{ntoinfty} n/sqrt[n] {n!} =e$ ($n$ integer) is a famous exercise and is already available on this website. I have given a similar argument replacing integer $n$ with real variable $x$.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:07










  • $begingroup$
    @DavidMB: I have added an update in my answer with some details.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:22














3












3








3





$begingroup$

First note that $x/(x!) ^{1/x}to e$ where by $x! $ we mean $Gamma (x+1)$ and hence the desired limit is equal to the limit of the expression $$ecdot(e^{1/x}-e^{sin(1/x)})x^2$$ Putting $x=1/t$ so that $tto 0^{+}$ we see that the desired limit is $$elim_{tto 0^{+}}e^{sin t} cdotfrac{e^{t-sin t} - 1}{t-sin t} cdot frac{t-sin t} {t^2}$$ Clearly the first two factors under limit operation tend to $1$ and the last one tends to $0$ and thus the limit in question is $0$.




The fact that $x/Gamma ^{1/x}(x+1)$ tends to $e$ as $xtoinfty $ can be established in the following manner.



Let $f(x) =log(x^x/Gamma (x+1))$ then $$f(x+1)-f(x)=xlogleft(1+frac {1}{x}right) $$ which tends to $1$ and hence $f(x) /xto 1$. And thus $log (x/Gamma ^{1/x}(x+1))to 1$.




The case when $x$ is an integer is famous and we can prove that $limlimits _{ntoinfty} dfrac{n} {sqrt[n] {n!}} =e$ using the well known




Theorem: If ${a_n} $ is a sequence of positive terms such that $a_{n+1}/a_nto L$ as $ntoinfty $ then $sqrt[n] {a_n} to L$ as $ntoinfty $.




Taking $a_n=n^n/n! $ we can see that $$frac{a_{n+1}}{a_n}=left(1+frac {1}{n}right) ^nto e$$ so that $sqrt[n] {a_n} =n/sqrt[n] {n!} to e$.



A similar theorem like the one above holds for functions of a real variable and the same has been used in my solution which uses $x$ as a real variable.






share|cite|improve this answer











$endgroup$



First note that $x/(x!) ^{1/x}to e$ where by $x! $ we mean $Gamma (x+1)$ and hence the desired limit is equal to the limit of the expression $$ecdot(e^{1/x}-e^{sin(1/x)})x^2$$ Putting $x=1/t$ so that $tto 0^{+}$ we see that the desired limit is $$elim_{tto 0^{+}}e^{sin t} cdotfrac{e^{t-sin t} - 1}{t-sin t} cdot frac{t-sin t} {t^2}$$ Clearly the first two factors under limit operation tend to $1$ and the last one tends to $0$ and thus the limit in question is $0$.




The fact that $x/Gamma ^{1/x}(x+1)$ tends to $e$ as $xtoinfty $ can be established in the following manner.



Let $f(x) =log(x^x/Gamma (x+1))$ then $$f(x+1)-f(x)=xlogleft(1+frac {1}{x}right) $$ which tends to $1$ and hence $f(x) /xto 1$. And thus $log (x/Gamma ^{1/x}(x+1))to 1$.




The case when $x$ is an integer is famous and we can prove that $limlimits _{ntoinfty} dfrac{n} {sqrt[n] {n!}} =e$ using the well known




Theorem: If ${a_n} $ is a sequence of positive terms such that $a_{n+1}/a_nto L$ as $ntoinfty $ then $sqrt[n] {a_n} to L$ as $ntoinfty $.




Taking $a_n=n^n/n! $ we can see that $$frac{a_{n+1}}{a_n}=left(1+frac {1}{n}right) ^nto e$$ so that $sqrt[n] {a_n} =n/sqrt[n] {n!} to e$.



A similar theorem like the one above holds for functions of a real variable and the same has been used in my solution which uses $x$ as a real variable.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 9:22

























answered Jan 3 at 5:03









Paramanand SinghParamanand Singh

49.7k556163




49.7k556163












  • $begingroup$
    Thank you for your solution. I am curious of how did you know it tends to e and what gave you the inclination to use substitution?
    $endgroup$
    – David MB
    Jan 3 at 6:58












  • $begingroup$
    @DavidMB: the limit $lim_{ntoinfty} n/sqrt[n] {n!} =e$ ($n$ integer) is a famous exercise and is already available on this website. I have given a similar argument replacing integer $n$ with real variable $x$.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:07










  • $begingroup$
    @DavidMB: I have added an update in my answer with some details.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:22


















  • $begingroup$
    Thank you for your solution. I am curious of how did you know it tends to e and what gave you the inclination to use substitution?
    $endgroup$
    – David MB
    Jan 3 at 6:58












  • $begingroup$
    @DavidMB: the limit $lim_{ntoinfty} n/sqrt[n] {n!} =e$ ($n$ integer) is a famous exercise and is already available on this website. I have given a similar argument replacing integer $n$ with real variable $x$.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:07










  • $begingroup$
    @DavidMB: I have added an update in my answer with some details.
    $endgroup$
    – Paramanand Singh
    Jan 3 at 9:22
















$begingroup$
Thank you for your solution. I am curious of how did you know it tends to e and what gave you the inclination to use substitution?
$endgroup$
– David MB
Jan 3 at 6:58






$begingroup$
Thank you for your solution. I am curious of how did you know it tends to e and what gave you the inclination to use substitution?
$endgroup$
– David MB
Jan 3 at 6:58














$begingroup$
@DavidMB: the limit $lim_{ntoinfty} n/sqrt[n] {n!} =e$ ($n$ integer) is a famous exercise and is already available on this website. I have given a similar argument replacing integer $n$ with real variable $x$.
$endgroup$
– Paramanand Singh
Jan 3 at 9:07




$begingroup$
@DavidMB: the limit $lim_{ntoinfty} n/sqrt[n] {n!} =e$ ($n$ integer) is a famous exercise and is already available on this website. I have given a similar argument replacing integer $n$ with real variable $x$.
$endgroup$
– Paramanand Singh
Jan 3 at 9:07












$begingroup$
@DavidMB: I have added an update in my answer with some details.
$endgroup$
– Paramanand Singh
Jan 3 at 9:22




$begingroup$
@DavidMB: I have added an update in my answer with some details.
$endgroup$
– Paramanand Singh
Jan 3 at 9:22


















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