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












2














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



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




















  • 1




    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
    – Andrew Shedlock
    yesterday










  • 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.
    – Matteo
    yesterday












  • We haven't learned the Gamma function, yet. Thank you though.
    – David MB
    yesterday
















2














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



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




















  • 1




    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
    – Andrew Shedlock
    yesterday










  • 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.
    – Matteo
    yesterday












  • We haven't learned the Gamma function, yet. Thank you though.
    – David MB
    yesterday














2












2








2


3





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



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















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



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 yesterday









zhw.

71.7k43075




71.7k43075










asked yesterday









David MB

267




267








  • 1




    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
    – Andrew Shedlock
    yesterday










  • 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.
    – Matteo
    yesterday












  • We haven't learned the Gamma function, yet. Thank you though.
    – David MB
    yesterday














  • 1




    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
    – Andrew Shedlock
    yesterday










  • 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.
    – Matteo
    yesterday












  • We haven't learned the Gamma function, yet. Thank you though.
    – David MB
    yesterday








1




1




Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
– Andrew Shedlock
yesterday




Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function.
– Andrew Shedlock
yesterday












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.
– Matteo
yesterday






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.
– Matteo
yesterday














We haven't learned the Gamma function, yet. Thank you though.
– David MB
yesterday




We haven't learned the Gamma function, yet. Thank you though.
– David MB
yesterday










2 Answers
2






active

oldest

votes


















1














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























  • Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    – David MB
    yesterday












  • I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    – Siong Thye Goh
    yesterday










  • I see. I'll definitely keep that in mind next time.
    – David MB
    yesterday



















3














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























  • 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?
    – David MB
    yesterday












  • @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 $x$.
    – Paramanand Singh
    yesterday










  • @DavidMB: I have added an update in my answer with some details.
    – Paramanand Singh
    yesterday











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060137%2ffind-lim-x-to-infty-left-frace1-x-e-sin1-xx3x1-x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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























  • Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    – David MB
    yesterday












  • I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    – Siong Thye Goh
    yesterday










  • I see. I'll definitely keep that in mind next time.
    – David MB
    yesterday
















1














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























  • Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    – David MB
    yesterday












  • I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    – Siong Thye Goh
    yesterday










  • I see. I'll definitely keep that in mind next time.
    – David MB
    yesterday














1












1








1






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














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 yesterday

























answered yesterday









Siong Thye Goh

99.5k1465117




99.5k1465117












  • Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    – David MB
    yesterday












  • I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    – Siong Thye Goh
    yesterday










  • I see. I'll definitely keep that in mind next time.
    – David MB
    yesterday


















  • Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
    – David MB
    yesterday












  • I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
    – Siong Thye Goh
    yesterday










  • I see. I'll definitely keep that in mind next time.
    – David MB
    yesterday
















Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
– David MB
yesterday






Many thanks. I was stuck on this for hours. Do you mind explaining where you got the fractions from (1/x)?
– David MB
yesterday














I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
– Siong Thye Goh
yesterday




I use taylor expansion of exponential and sine functions. keeping track of the dominant terms.
– Siong Thye Goh
yesterday












I see. I'll definitely keep that in mind next time.
– David MB
yesterday




I see. I'll definitely keep that in mind next time.
– David MB
yesterday











3














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























  • 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?
    – David MB
    yesterday












  • @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 $x$.
    – Paramanand Singh
    yesterday










  • @DavidMB: I have added an update in my answer with some details.
    – Paramanand Singh
    yesterday
















3














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























  • 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?
    – David MB
    yesterday












  • @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 $x$.
    – Paramanand Singh
    yesterday










  • @DavidMB: I have added an update in my answer with some details.
    – Paramanand Singh
    yesterday














3












3








3






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














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 yesterday

























answered yesterday









Paramanand Singh

49k555160




49k555160












  • 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?
    – David MB
    yesterday












  • @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 $x$.
    – Paramanand Singh
    yesterday










  • @DavidMB: I have added an update in my answer with some details.
    – Paramanand Singh
    yesterday


















  • 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?
    – David MB
    yesterday












  • @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 $x$.
    – Paramanand Singh
    yesterday










  • @DavidMB: I have added an update in my answer with some details.
    – Paramanand Singh
    yesterday
















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?
– David MB
yesterday






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?
– David MB
yesterday














@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 $x$.
– Paramanand Singh
yesterday




@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 $x$.
– Paramanand Singh
yesterday












@DavidMB: I have added an update in my answer with some details.
– Paramanand Singh
yesterday




@DavidMB: I have added an update in my answer with some details.
– Paramanand Singh
yesterday


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060137%2ffind-lim-x-to-infty-left-frace1-x-e-sin1-xx3x1-x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

An IMO inspired problem

Image
14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
Read more

Management

Image
Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
Read more

Has there ever been an instance of an active nuclear power plant within or near a war zone?

Image
36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
Read more