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How to calculate the limit $lim_xtoinfty (x^1/n-ln(x))$
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As in the title, I am having trouble with this seemingly easy limit. I cannot seem to transform it into a form that would let us use L'Hôpital. I also cannot see anything I can multiply by to simplify the limit. I was wondering if we can perhaps use the fact that
$$
lim_xtoinftyfracln(x)x^frac1n=0.
$$
Or maybe raise the limit to the $n$th power but that would require a messy binomial. I am probably missing something obvious but I would appreciate any hints.
limits
edited Jul 27 at 14:30
egreg
164k1180187
asked Jul 27 at 14:25
Sorfosh
910616
add a comment |Â
up vote
1
down vote
favorite
As in the title, I am having trouble with this seemingly easy limit. I cannot seem to transform it into a form that would let us use L'Hôpital. I also cannot see anything I can multiply by to simplify the limit. I was wondering if we can perhaps use the fact that
$$
lim_xtoinftyfracln(x)x^frac1n=0.
$$
Or maybe raise the limit to the $n$th power but that would require a messy binomial. I am probably missing something obvious but I would appreciate any hints.
limits
edited Jul 27 at 14:30
egreg
164k1180187
asked Jul 27 at 14:25
Sorfosh
910616
$limlimits_xtoinftyfracln(x)x^frac1n=0$ and $limlimits_xtoinftyx^frac1n=+infty$ together imply $limlimits_xtoinfty x^1/n-ln(x) = +infty$
– Henry
Jul 27 at 14:29
You can get rid of the root by the substitution $z:=x^1/n$ and the limits turns to $lim_ztoinfty(z-nln z)=lim_ztoinftyz(1-nln z/z)$.
– Yves Daoust
Jul 27 at 14:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
As in the title, I am having trouble with this seemingly easy limit. I cannot seem to transform it into a form that would let us use L'Hôpital. I also cannot see anything I can multiply by to simplify the limit. I was wondering if we can perhaps use the fact that
$$
lim_xtoinftyfracln(x)x^frac1n=0.
$$
Or maybe raise the limit to the $n$th power but that would require a messy binomial. I am probably missing something obvious but I would appreciate any hints.
limits
edited Jul 27 at 14:30
egreg
164k1180187
asked Jul 27 at 14:25
Sorfosh
910616
As in the title, I am having trouble with this seemingly easy limit. I cannot seem to transform it into a form that would let us use L'Hôpital. I also cannot see anything I can multiply by to simplify the limit. I was wondering if we can perhaps use the fact that
$$
lim_xtoinftyfracln(x)x^frac1n=0.
$$
Or maybe raise the limit to the $n$th power but that would require a messy binomial. I am probably missing something obvious but I would appreciate any hints.
limits
edited Jul 27 at 14:30
egreg
164k1180187
asked Jul 27 at 14:25
Sorfosh
910616
edited Jul 27 at 14:30
egreg
164k1180187
edited Jul 27 at 14:30
egreg
164k1180187
edited Jul 27 at 14:30
egreg
164k1180187
164k1180187
asked Jul 27 at 14:25
Sorfosh
910616
asked Jul 27 at 14:25
Sorfosh
910616
asked Jul 27 at 14:25
Sorfosh
910616
910616
$limlimits_xtoinftyfracln(x)x^frac1n=0$ and $limlimits_xtoinftyx^frac1n=+infty$ together imply $limlimits_xtoinfty x^1/n-ln(x) = +infty$
– Henry
Jul 27 at 14:29
You can get rid of the root by the substitution $z:=x^1/n$ and the limits turns to $lim_ztoinfty(z-nln z)=lim_ztoinftyz(1-nln z/z)$.
– Yves Daoust
Jul 27 at 14:32
add a comment |Â
$limlimits_xtoinftyfracln(x)x^frac1n=0$ and $limlimits_xtoinftyx^frac1n=+infty$ together imply $limlimits_xtoinfty x^1/n-ln(x) = +infty$
– Henry
Jul 27 at 14:29
You can get rid of the root by the substitution $z:=x^1/n$ and the limits turns to $lim_ztoinfty(z-nln z)=lim_ztoinftyz(1-nln z/z)$.
– Yves Daoust
Jul 27 at 14:32
$limlimits_xtoinftyfracln(x)x^frac1n=0$ and $limlimits_xtoinftyx^frac1n=+infty$ together imply $limlimits_xtoinfty x^1/n-ln(x) = +infty$
– Henry
Jul 27 at 14:29
$limlimits_xtoinftyfracln(x)x^frac1n=0$ and $limlimits_xtoinftyx^frac1n=+infty$ together imply $limlimits_xtoinfty x^1/n-ln(x) = +infty$
– Henry
Jul 27 at 14:29
You can get rid of the root by the substitution $z:=x^1/n$ and the limits turns to $lim_ztoinfty(z-nln z)=lim_ztoinftyz(1-nln z/z)$.
– Yves Daoust
Jul 27 at 14:32
You can get rid of the root by the substitution $z:=x^1/n$ and the limits turns to $lim_ztoinfty(z-nln z)=lim_ztoinftyz(1-nln z/z)$.
– Yves Daoust
Jul 27 at 14:32
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
Yes, you can use that fact:
$$
lim_xtoinfty(x^1/n-ln x)=
lim_xtoinftyleft(fracx^1/nln x-1right)ln x
$$
The limit of the part in parentheses is $infty$.
answered Jul 27 at 14:32
egreg
164k1180187
add a comment |Â
up vote
1
down vote
Hint:
Set $t=x^tfrac1n$, so the expression can be re-written as
$$x^tfrac1n-ln x=t-nln t$=tBigl(1-n,fracln ttBigr),$$
and use that near $infty$, $:ln t=o(t)$.
answered Jul 27 at 14:35
Bernard
110k635102
add a comment |Â
up vote
0
down vote
We have
$$x^1/n-ln(x)=ln xleft(fracx^1/nln x-1right)to infty$$
indeed $ln x to infty$ and for $x=e^y$ with $y to infty$
$$fracx^1/nln x=frace^y/ny=frac1nfrace^y/ny/nto infty$$
indeed eventually as $t to infty$ we have $e^t>t^2$ and then
$$frace^y/ny/nge frac(y/n)^2y/n=frac y n to infty$$
answered Jul 27 at 14:34
gimusi
64.9k73583
add a comment |Â
up vote
0
down vote
Use that:
$$lim_nto k [f(n)-g(n)]=lim_nto k[f(n)]-lim_nto k[g(n)]$$
answered Jul 27 at 14:57
Rhys Hughes
3,8581227
That's just $infty-infty$
– Sorfosh
Jul 27 at 15:15
The limit is in respect to $x$, not $n$
– Sorfosh
Jul 27 at 15:18
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, you can use that fact:
$$
lim_xtoinfty(x^1/n-ln x)=
lim_xtoinftyleft(fracx^1/nln x-1right)ln x
$$
The limit of the part in parentheses is $infty$.
answered Jul 27 at 14:32
egreg
164k1180187
add a comment |Â
up vote
2
down vote
accepted
Yes, you can use that fact:
$$
lim_xtoinfty(x^1/n-ln x)=
lim_xtoinftyleft(fracx^1/nln x-1right)ln x
$$
The limit of the part in parentheses is $infty$.
answered Jul 27 at 14:32
egreg
164k1180187
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, you can use that fact:
$$
lim_xtoinfty(x^1/n-ln x)=
lim_xtoinftyleft(fracx^1/nln x-1right)ln x
$$
The limit of the part in parentheses is $infty$.
answered Jul 27 at 14:32
egreg
164k1180187
Yes, you can use that fact:
$$
lim_xtoinfty(x^1/n-ln x)=
lim_xtoinftyleft(fracx^1/nln x-1right)ln x
$$
The limit of the part in parentheses is $infty$.
answered Jul 27 at 14:32
egreg
164k1180187
answered Jul 27 at 14:32
egreg
164k1180187
answered Jul 27 at 14:32
egreg
164k1180187
answered Jul 27 at 14:32
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint:
Set $t=x^tfrac1n$, so the expression can be re-written as
$$x^tfrac1n-ln x=t-nln t$=tBigl(1-n,fracln ttBigr),$$
and use that near $infty$, $:ln t=o(t)$.
answered Jul 27 at 14:35
Bernard
110k635102
add a comment |Â
up vote
1
down vote
Hint:
Set $t=x^tfrac1n$, so the expression can be re-written as
$$x^tfrac1n-ln x=t-nln t$=tBigl(1-n,fracln ttBigr),$$
and use that near $infty$, $:ln t=o(t)$.
answered Jul 27 at 14:35
Bernard
110k635102
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint:
Set $t=x^tfrac1n$, so the expression can be re-written as
$$x^tfrac1n-ln x=t-nln t$=tBigl(1-n,fracln ttBigr),$$
and use that near $infty$, $:ln t=o(t)$.
answered Jul 27 at 14:35
Bernard
110k635102
Hint:
Set $t=x^tfrac1n$, so the expression can be re-written as
$$x^tfrac1n-ln x=t-nln t$=tBigl(1-n,fracln ttBigr),$$
and use that near $infty$, $:ln t=o(t)$.
answered Jul 27 at 14:35
Bernard
110k635102
answered Jul 27 at 14:35
Bernard
110k635102
answered Jul 27 at 14:35
Bernard
110k635102
answered Jul 27 at 14:35
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
up vote
0
down vote
We have
$$x^1/n-ln(x)=ln xleft(fracx^1/nln x-1right)to infty$$
indeed $ln x to infty$ and for $x=e^y$ with $y to infty$
$$fracx^1/nln x=frace^y/ny=frac1nfrace^y/ny/nto infty$$
indeed eventually as $t to infty$ we have $e^t>t^2$ and then
$$frace^y/ny/nge frac(y/n)^2y/n=frac y n to infty$$
answered Jul 27 at 14:34
gimusi
64.9k73583
add a comment |Â
up vote
0
down vote
We have
$$x^1/n-ln(x)=ln xleft(fracx^1/nln x-1right)to infty$$
indeed $ln x to infty$ and for $x=e^y$ with $y to infty$
$$fracx^1/nln x=frace^y/ny=frac1nfrace^y/ny/nto infty$$
indeed eventually as $t to infty$ we have $e^t>t^2$ and then
$$frace^y/ny/nge frac(y/n)^2y/n=frac y n to infty$$
answered Jul 27 at 14:34
gimusi
64.9k73583
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have
$$x^1/n-ln(x)=ln xleft(fracx^1/nln x-1right)to infty$$
indeed $ln x to infty$ and for $x=e^y$ with $y to infty$
$$fracx^1/nln x=frace^y/ny=frac1nfrace^y/ny/nto infty$$
indeed eventually as $t to infty$ we have $e^t>t^2$ and then
$$frace^y/ny/nge frac(y/n)^2y/n=frac y n to infty$$
answered Jul 27 at 14:34
gimusi
64.9k73583
We have
$$x^1/n-ln(x)=ln xleft(fracx^1/nln x-1right)to infty$$
indeed $ln x to infty$ and for $x=e^y$ with $y to infty$
$$fracx^1/nln x=frace^y/ny=frac1nfrace^y/ny/nto infty$$
indeed eventually as $t to infty$ we have $e^t>t^2$ and then
$$frace^y/ny/nge frac(y/n)^2y/n=frac y n to infty$$
answered Jul 27 at 14:34
gimusi
64.9k73583
answered Jul 27 at 14:34
gimusi
64.9k73583
answered Jul 27 at 14:34
gimusi
64.9k73583
answered Jul 27 at 14:34
gimusi
64.9k73583
64.9k73583
add a comment |Â
add a comment |Â
up vote
0
down vote
Use that:
$$lim_nto k [f(n)-g(n)]=lim_nto k[f(n)]-lim_nto k[g(n)]$$
answered Jul 27 at 14:57
Rhys Hughes
3,8581227
That's just $infty-infty$
– Sorfosh
Jul 27 at 15:15
The limit is in respect to $x$, not $n$
– Sorfosh
Jul 27 at 15:18
add a comment |Â
up vote
0
down vote
Use that:
$$lim_nto k [f(n)-g(n)]=lim_nto k[f(n)]-lim_nto k[g(n)]$$
answered Jul 27 at 14:57
Rhys Hughes
3,8581227
That's just $infty-infty$
– Sorfosh
Jul 27 at 15:15
The limit is in respect to $x$, not $n$
– Sorfosh
Jul 27 at 15:18
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use that:
$$lim_nto k [f(n)-g(n)]=lim_nto k[f(n)]-lim_nto k[g(n)]$$
answered Jul 27 at 14:57
Rhys Hughes
3,8581227
Use that:
$$lim_nto k [f(n)-g(n)]=lim_nto k[f(n)]-lim_nto k[g(n)]$$
answered Jul 27 at 14:57
Rhys Hughes
3,8581227
answered Jul 27 at 14:57
Rhys Hughes
3,8581227
answered Jul 27 at 14:57
Rhys Hughes
3,8581227
answered Jul 27 at 14:57
Rhys Hughes
3,8581227
3,8581227
That's just $infty-infty$
– Sorfosh
Jul 27 at 15:15
The limit is in respect to $x$, not $n$
– Sorfosh
Jul 27 at 15:18
add a comment |Â
That's just $infty-infty$
– Sorfosh
Jul 27 at 15:15
The limit is in respect to $x$, not $n$
– Sorfosh
Jul 27 at 15:18
That's just $infty-infty$
– Sorfosh
Jul 27 at 15:15
That's just $infty-infty$
– Sorfosh
Jul 27 at 15:15
The limit is in respect to $x$, not $n$
– Sorfosh
Jul 27 at 15:18
The limit is in respect to $x$, not $n$
– Sorfosh
Jul 27 at 15:18
add a comment |Â
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$limlimits_xtoinftyfracln(x)x^frac1n=0$ and $limlimits_xtoinftyx^frac1n=+infty$ together imply $limlimits_xtoinfty x^1/n-ln(x) = +infty$
– Henry
Jul 27 at 14:29
You can get rid of the root by the substitution $z:=x^1/n$ and the limits turns to $lim_ztoinfty(z-nln z)=lim_ztoinftyz(1-nln z/z)$.
– Yves Daoust
Jul 27 at 14:32