Limit with integral and powerHow can I calculate the following limit? $lim_ntoinftyfrac1(ln n)^ln n$Calculate a limit where $sin n_k$ increases monotonically to 1.Limit of a Riemann Sum and IntegralLimit with Integral and Sigmahelp on limit exerciseHow to solve this integral and limit?prove limit of $ n^(1/loglog n) /bigl( ((log n)^cbigr)$Limit of an integralCurious limit with variable under integral?Evaluating $limlimits_xtoinftyxint_0^pi/4ln(1+tan^xt)dt$
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Limit with integral and power
How can I calculate the following limit? $lim_ntoinftyfrac1(ln n)^ln n$Calculate a limit where $sin n_k$ increases monotonically to 1.Limit of a Riemann Sum and IntegralLimit with Integral and Sigmahelp on limit exerciseHow to solve this integral and limit?prove limit of $ n^(1/loglog n) /bigl( ((log n)^cbigr)$Limit of an integralCurious limit with variable under integral?Evaluating $limlimits_xtoinftyxint_0^pi/4ln(1+tan^xt)dt$
$begingroup$
I'm trying to calculate this limit:
$$lim_xtoinfty left(int_0^1 t^-tx dtright)^frac1x$$
I tried the squeezing idea without success.
calculus limits definite-integrals
edited Mar 17 at 23:35
RRL
53k42573
asked Feb 23 '17 at 19:13
song01song01
613
$endgroup$
add a comment |
$begingroup$
I'm trying to calculate this limit:
$$lim_xtoinfty left(int_0^1 t^-tx dtright)^frac1x$$
I tried the squeezing idea without success.
calculus limits definite-integrals
edited Mar 17 at 23:35
RRL
53k42573
asked Feb 23 '17 at 19:13
song01song01
613
$endgroup$
2
$begingroup$
If $f$ is continuous or just integrable , $lim_n to infty left(int_0^1 |f(x)|^n, dxright)^1/n = sup_x in [0,1] f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
add a comment |
$begingroup$
I'm trying to calculate this limit:
$$lim_xtoinfty left(int_0^1 t^-tx dtright)^frac1x$$
I tried the squeezing idea without success.
calculus limits definite-integrals
edited Mar 17 at 23:35
RRL
53k42573
asked Feb 23 '17 at 19:13
song01song01
613
$endgroup$
I'm trying to calculate this limit:
$$lim_xtoinfty left(int_0^1 t^-tx dtright)^frac1x$$
I tried the squeezing idea without success.
calculus limits definite-integrals
calculus limits definite-integrals
edited Mar 17 at 23:35
RRL
53k42573
asked Feb 23 '17 at 19:13
song01song01
613
edited Mar 17 at 23:35
RRL
53k42573
asked Feb 23 '17 at 19:13
song01song01
613
edited Mar 17 at 23:35
RRL
53k42573
edited Mar 17 at 23:35
RRL
53k42573
edited Mar 17 at 23:35
RRL
53k42573
53k42573
asked Feb 23 '17 at 19:13
song01song01
613
asked Feb 23 '17 at 19:13
song01song01
613
asked Feb 23 '17 at 19:13
song01song01
613
613
2
$begingroup$
If $f$ is continuous or just integrable , $lim_n to infty left(int_0^1 |f(x)|^n, dxright)^1/n = sup_x in [0,1] f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
add a comment |
2
$begingroup$
If $f$ is continuous or just integrable , $lim_n to infty left(int_0^1 |f(x)|^n, dxright)^1/n = sup_x in [0,1] f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
2
2
$begingroup$
If $f$ is continuous or just integrable , $lim_n to infty left(int_0^1 |f(x)|^n, dxright)^1/n = sup_x in [0,1] f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
$begingroup$
If $f$ is continuous or just integrable , $lim_n to infty left(int_0^1 |f(x)|^n, dxright)^1/n = sup_x in [0,1] f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For any large $x$ we have
$$ int_0^1t^-tx,dt = int_0^1expleft(-x tlog tright),dt = sum_ngeq 0fracx^nn!int_0^1t^n(-log t)^n,dt =sum_ngeq 0fracx^nn!(n+1)^n+1$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_xto +inftyfraclog f(x)x stackreldH= explim_xto +inftyfracf'(x)f(x)=explim_xto +inftyfracint_0^1(-tlog t)t^-tx,dtint_0^1t^-tx,dt$$
that is $colorrede^1/e$ since $t^-tx$ converges in distribution to $Ccdotdeltaleft(t-frac1eright)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
A maximum for $t^-t$ on $[0,1]$ is attained at $t = e^-1.$
With $0 < t < 1,$ we have
$$t^-tx = exp(-t ln t)^x leqslant exp(e^-1)^x \ implies left(int_0^1 t^-tx , dtright)^1/x leqslant exp(e^-1).$$
Using continuity, for any small $epsilon$, there is an interval $[e^-1 - delta, e^-1 + delta] subset [0,1]$ where $t^-t > exp(e^-1) - epsilon$ and
$$left(int_0^1 t^-tx , dtright)^1/x geqslant left(int_exp(e^-1) - delta^exp(e^-1) + delta t^-tx , dtright)^1/x > (exp(e^-1)- epsilon)(2 delta)^1/x. $$
Since $(2delta)^1/x to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_x to inftyleft(int_0^1 t^-tx , dtright)^1/x = exp(e^-1)$$
answered Feb 23 '17 at 19:25
RRLRRL
53k42573
$endgroup$
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any large $x$ we have
$$ int_0^1t^-tx,dt = int_0^1expleft(-x tlog tright),dt = sum_ngeq 0fracx^nn!int_0^1t^n(-log t)^n,dt =sum_ngeq 0fracx^nn!(n+1)^n+1$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_xto +inftyfraclog f(x)x stackreldH= explim_xto +inftyfracf'(x)f(x)=explim_xto +inftyfracint_0^1(-tlog t)t^-tx,dtint_0^1t^-tx,dt$$
that is $colorrede^1/e$ since $t^-tx$ converges in distribution to $Ccdotdeltaleft(t-frac1eright)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
For any large $x$ we have
$$ int_0^1t^-tx,dt = int_0^1expleft(-x tlog tright),dt = sum_ngeq 0fracx^nn!int_0^1t^n(-log t)^n,dt =sum_ngeq 0fracx^nn!(n+1)^n+1$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_xto +inftyfraclog f(x)x stackreldH= explim_xto +inftyfracf'(x)f(x)=explim_xto +inftyfracint_0^1(-tlog t)t^-tx,dtint_0^1t^-tx,dt$$
that is $colorrede^1/e$ since $t^-tx$ converges in distribution to $Ccdotdeltaleft(t-frac1eright)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
For any large $x$ we have
$$ int_0^1t^-tx,dt = int_0^1expleft(-x tlog tright),dt = sum_ngeq 0fracx^nn!int_0^1t^n(-log t)^n,dt =sum_ngeq 0fracx^nn!(n+1)^n+1$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_xto +inftyfraclog f(x)x stackreldH= explim_xto +inftyfracf'(x)f(x)=explim_xto +inftyfracint_0^1(-tlog t)t^-tx,dtint_0^1t^-tx,dt$$
that is $colorrede^1/e$ since $t^-tx$ converges in distribution to $Ccdotdeltaleft(t-frac1eright)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
For any large $x$ we have
$$ int_0^1t^-tx,dt = int_0^1expleft(-x tlog tright),dt = sum_ngeq 0fracx^nn!int_0^1t^n(-log t)^n,dt =sum_ngeq 0fracx^nn!(n+1)^n+1$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_xto +inftyfraclog f(x)x stackreldH= explim_xto +inftyfracf'(x)f(x)=explim_xto +inftyfracint_0^1(-tlog t)t^-tx,dtint_0^1t^-tx,dt$$
that is $colorrede^1/e$ since $t^-tx$ converges in distribution to $Ccdotdeltaleft(t-frac1eright)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
292k33284672
add a comment |
add a comment |
$begingroup$
A maximum for $t^-t$ on $[0,1]$ is attained at $t = e^-1.$
With $0 < t < 1,$ we have
$$t^-tx = exp(-t ln t)^x leqslant exp(e^-1)^x \ implies left(int_0^1 t^-tx , dtright)^1/x leqslant exp(e^-1).$$
Using continuity, for any small $epsilon$, there is an interval $[e^-1 - delta, e^-1 + delta] subset [0,1]$ where $t^-t > exp(e^-1) - epsilon$ and
$$left(int_0^1 t^-tx , dtright)^1/x geqslant left(int_exp(e^-1) - delta^exp(e^-1) + delta t^-tx , dtright)^1/x > (exp(e^-1)- epsilon)(2 delta)^1/x. $$
Since $(2delta)^1/x to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_x to inftyleft(int_0^1 t^-tx , dtright)^1/x = exp(e^-1)$$
answered Feb 23 '17 at 19:25
RRLRRL
53k42573
$endgroup$
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
$begingroup$
A maximum for $t^-t$ on $[0,1]$ is attained at $t = e^-1.$
With $0 < t < 1,$ we have
$$t^-tx = exp(-t ln t)^x leqslant exp(e^-1)^x \ implies left(int_0^1 t^-tx , dtright)^1/x leqslant exp(e^-1).$$
Using continuity, for any small $epsilon$, there is an interval $[e^-1 - delta, e^-1 + delta] subset [0,1]$ where $t^-t > exp(e^-1) - epsilon$ and
$$left(int_0^1 t^-tx , dtright)^1/x geqslant left(int_exp(e^-1) - delta^exp(e^-1) + delta t^-tx , dtright)^1/x > (exp(e^-1)- epsilon)(2 delta)^1/x. $$
Since $(2delta)^1/x to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_x to inftyleft(int_0^1 t^-tx , dtright)^1/x = exp(e^-1)$$
answered Feb 23 '17 at 19:25
RRLRRL
53k42573
$endgroup$
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
$begingroup$
A maximum for $t^-t$ on $[0,1]$ is attained at $t = e^-1.$
With $0 < t < 1,$ we have
$$t^-tx = exp(-t ln t)^x leqslant exp(e^-1)^x \ implies left(int_0^1 t^-tx , dtright)^1/x leqslant exp(e^-1).$$
Using continuity, for any small $epsilon$, there is an interval $[e^-1 - delta, e^-1 + delta] subset [0,1]$ where $t^-t > exp(e^-1) - epsilon$ and
$$left(int_0^1 t^-tx , dtright)^1/x geqslant left(int_exp(e^-1) - delta^exp(e^-1) + delta t^-tx , dtright)^1/x > (exp(e^-1)- epsilon)(2 delta)^1/x. $$
Since $(2delta)^1/x to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_x to inftyleft(int_0^1 t^-tx , dtright)^1/x = exp(e^-1)$$
answered Feb 23 '17 at 19:25
RRLRRL
53k42573
$endgroup$
A maximum for $t^-t$ on $[0,1]$ is attained at $t = e^-1.$
With $0 < t < 1,$ we have
$$t^-tx = exp(-t ln t)^x leqslant exp(e^-1)^x \ implies left(int_0^1 t^-tx , dtright)^1/x leqslant exp(e^-1).$$
Using continuity, for any small $epsilon$, there is an interval $[e^-1 - delta, e^-1 + delta] subset [0,1]$ where $t^-t > exp(e^-1) - epsilon$ and
$$left(int_0^1 t^-tx , dtright)^1/x geqslant left(int_exp(e^-1) - delta^exp(e^-1) + delta t^-tx , dtright)^1/x > (exp(e^-1)- epsilon)(2 delta)^1/x. $$
Since $(2delta)^1/x to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_x to inftyleft(int_0^1 t^-tx , dtright)^1/x = exp(e^-1)$$
answered Feb 23 '17 at 19:25
RRLRRL
53k42573
edited Feb 23 '17 at 19:46
answered Feb 23 '17 at 19:25
RRLRRL
53k42573
answered Feb 23 '17 at 19:25
RRLRRL
53k42573
answered Feb 23 '17 at 19:25
RRLRRL
53k42573
53k42573
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
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$begingroup$
If $f$ is continuous or just integrable , $lim_n to infty left(int_0^1 |f(x)|^n, dxright)^1/n = sup_x in [0,1] f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30