Find $lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{{(x)}^3}}{x!^{(1/x)}}right)$
$$lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}right)$$
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
add a comment |
$$lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}right)$$
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
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
add a comment |
$$lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}right)$$
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
$$lim_{xtoinfty}left(frac{({e^{1/x}-e^{sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}right)$$
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
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
add a comment |
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
add a comment |
2 Answers
2
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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}
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
add a comment |
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.
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
add a comment |
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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}
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
add a comment |
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}
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
add a comment |
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}
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}
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
add a comment |
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
add a comment |
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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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