Mixing
limit of and integral depending on n
$begingroup$
Can somebody give me some tips about how can find the next limit, please?
$$ lim_{nto infty} int_{0}^4 sqrt[n]{x^n+(4-x)^n} dx $$I found that $$int_{0}^4 sqrt[n]{x^n+(4-x)^n} =2int_{0}^2 sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.
calculus limits definite-integrals
asked Jan 7 at 14:33
GaboruGaboru
637
$endgroup$
add a comment |
$begingroup$
Can somebody give me some tips about how can find the next limit, please?
$$ lim_{nto infty} int_{0}^4 sqrt[n]{x^n+(4-x)^n} dx $$I found that $$int_{0}^4 sqrt[n]{x^n+(4-x)^n} =2int_{0}^2 sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.
calculus limits definite-integrals
asked Jan 7 at 14:33
GaboruGaboru
637
$endgroup$
$begingroup$
I guess it should be $ntoinfty$, not $x$.
$endgroup$
– Mark
Jan 7 at 14:34
1
$begingroup$
Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
$endgroup$
– Math_QED
Jan 7 at 14:56
$begingroup$
@Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
$endgroup$
– Mark
Jan 7 at 15:18
add a comment |
$begingroup$
Can somebody give me some tips about how can find the next limit, please?
$$ lim_{nto infty} int_{0}^4 sqrt[n]{x^n+(4-x)^n} dx $$I found that $$int_{0}^4 sqrt[n]{x^n+(4-x)^n} =2int_{0}^2 sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.
calculus limits definite-integrals
asked Jan 7 at 14:33
GaboruGaboru
637
$endgroup$
Can somebody give me some tips about how can find the next limit, please?
$$ lim_{nto infty} int_{0}^4 sqrt[n]{x^n+(4-x)^n} dx $$I found that $$int_{0}^4 sqrt[n]{x^n+(4-x)^n} =2int_{0}^2 sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.
calculus limits definite-integrals
calculus limits definite-integrals
asked Jan 7 at 14:33
GaboruGaboru
637
asked Jan 7 at 14:33
GaboruGaboru
637
edited Jan 7 at 14:45
Gaboru
asked Jan 7 at 14:33
GaboruGaboru
637
asked Jan 7 at 14:33
GaboruGaboru
637
asked Jan 7 at 14:33
GaboruGaboru
637
637
$begingroup$
I guess it should be $ntoinfty$, not $x$.
$endgroup$
– Mark
Jan 7 at 14:34
1
$begingroup$
Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
$endgroup$
– Math_QED
Jan 7 at 14:56
$begingroup$
@Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
$endgroup$
– Mark
Jan 7 at 15:18
add a comment |
$begingroup$
I guess it should be $ntoinfty$, not $x$.
$endgroup$
– Mark
Jan 7 at 14:34
1
$begingroup$
Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
$endgroup$
– Math_QED
Jan 7 at 14:56
$begingroup$
@Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
$endgroup$
– Mark
Jan 7 at 15:18
$begingroup$
I guess it should be $ntoinfty$, not $x$.
$endgroup$
– Mark
Jan 7 at 14:34
$begingroup$
I guess it should be $ntoinfty$, not $x$.
$endgroup$
– Mark
Jan 7 at 14:34
1
1
$begingroup$
Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
$endgroup$
– Math_QED
Jan 7 at 14:56
$begingroup$
Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
$endgroup$
– Math_QED
Jan 7 at 14:56
$begingroup$
@Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
$endgroup$
– Mark
Jan 7 at 15:18
$begingroup$
@Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
$endgroup$
– Mark
Jan 7 at 15:18
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that the p-norm
of a 2D vector is defined as
$$
left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
$$
and it is known that
$$
mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
$$
Therefore
$$
eqalign{
& mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
& = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
& = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
& = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
$$
answered Jan 9 at 18:40
G CabG Cab
18.7k31238
$endgroup$
add a comment |
$begingroup$
Hint: assuming Mark’s comment is correct,
$2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$
answered Jan 7 at 14:40
MindlackMindlack
3,27717
$endgroup$
$begingroup$
Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
$endgroup$
– Gaboru
Jan 7 at 14:47
$begingroup$
Take everything to the $n$-th powe, maybe?
$endgroup$
– Mindlack
Jan 7 at 15:16
$begingroup$
I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
$endgroup$
– DavidG
Jan 9 at 4:35
add a comment |
$begingroup$
NOT A SOLUTION:
Not sure if this will be of help. I would start by repositioning the integrand:
begin{equation}
I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
end{equation}
Now let $t = dfrac{x}{4 - x}$ :
begin{equation}
I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
end{equation}
I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.
begin{equation}
lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
end{equation}
Then,
begin{align}
I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
end{align}
But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.
answered Jan 8 at 12:55
DavidGDavidG
2,1671720
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that the p-norm
of a 2D vector is defined as
$$
left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
$$
and it is known that
$$
mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
$$
Therefore
$$
eqalign{
& mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
& = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
& = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
& = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
$$
answered Jan 9 at 18:40
G CabG Cab
18.7k31238
$endgroup$
add a comment |
$begingroup$
Note that the p-norm
of a 2D vector is defined as
$$
left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
$$
and it is known that
$$
mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
$$
Therefore
$$
eqalign{
& mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
& = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
& = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
& = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
$$
answered Jan 9 at 18:40
G CabG Cab
18.7k31238
$endgroup$
add a comment |
$begingroup$
Note that the p-norm
of a 2D vector is defined as
$$
left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
$$
and it is known that
$$
mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
$$
Therefore
$$
eqalign{
& mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
& = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
& = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
& = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
$$
answered Jan 9 at 18:40
G CabG Cab
18.7k31238
$endgroup$
Note that the p-norm
of a 2D vector is defined as
$$
left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
$$
and it is known that
$$
mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
$$
Therefore
$$
eqalign{
& mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
& = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
& = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
& = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
$$
answered Jan 9 at 18:40
G CabG Cab
18.7k31238
answered Jan 9 at 18:40
G CabG Cab
18.7k31238
answered Jan 9 at 18:40
G CabG Cab
18.7k31238
answered Jan 9 at 18:40
G CabG Cab
18.7k31238
18.7k31238
add a comment |
add a comment |
$begingroup$
Hint: assuming Mark’s comment is correct,
$2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$
answered Jan 7 at 14:40
MindlackMindlack
3,27717
$endgroup$
$begingroup$
Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
$endgroup$
– Gaboru
Jan 7 at 14:47
$begingroup$
Take everything to the $n$-th powe, maybe?
$endgroup$
– Mindlack
Jan 7 at 15:16
$begingroup$
I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
$endgroup$
– DavidG
Jan 9 at 4:35
add a comment |
$begingroup$
Hint: assuming Mark’s comment is correct,
$2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$
answered Jan 7 at 14:40
MindlackMindlack
3,27717
$endgroup$
$begingroup$
Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
$endgroup$
– Gaboru
Jan 7 at 14:47
$begingroup$
Take everything to the $n$-th powe, maybe?
$endgroup$
– Mindlack
Jan 7 at 15:16
$begingroup$
I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
$endgroup$
– DavidG
Jan 9 at 4:35
add a comment |
$begingroup$
Hint: assuming Mark’s comment is correct,
$2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$
answered Jan 7 at 14:40
MindlackMindlack
3,27717
$endgroup$
Hint: assuming Mark’s comment is correct,
$2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$
answered Jan 7 at 14:40
MindlackMindlack
3,27717
answered Jan 7 at 14:40
MindlackMindlack
3,27717
answered Jan 7 at 14:40
MindlackMindlack
3,27717
answered Jan 7 at 14:40
MindlackMindlack
3,27717
3,27717
$begingroup$
Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
$endgroup$
– Gaboru
Jan 7 at 14:47
$begingroup$
Take everything to the $n$-th powe, maybe?
$endgroup$
– Mindlack
Jan 7 at 15:16
$begingroup$
I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
$endgroup$
– DavidG
Jan 9 at 4:35
add a comment |
$begingroup$
Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
$endgroup$
– Gaboru
Jan 7 at 14:47
$begingroup$
Take everything to the $n$-th powe, maybe?
$endgroup$
– Mindlack
Jan 7 at 15:16
$begingroup$
I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
$endgroup$
– DavidG
Jan 9 at 4:35
$begingroup$
Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
$endgroup$
– Gaboru
Jan 7 at 14:47
$begingroup$
Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
$endgroup$
– Gaboru
Jan 7 at 14:47
$begingroup$
Take everything to the $n$-th powe, maybe?
$endgroup$
– Mindlack
Jan 7 at 15:16
$begingroup$
Take everything to the $n$-th powe, maybe?
$endgroup$
– Mindlack
Jan 7 at 15:16
$begingroup$
I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
$endgroup$
– DavidG
Jan 9 at 4:35
$begingroup$
I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
$endgroup$
– DavidG
Jan 9 at 4:35
add a comment |
$begingroup$
NOT A SOLUTION:
Not sure if this will be of help. I would start by repositioning the integrand:
begin{equation}
I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
end{equation}
Now let $t = dfrac{x}{4 - x}$ :
begin{equation}
I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
end{equation}
I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.
begin{equation}
lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
end{equation}
Then,
begin{align}
I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
end{align}
But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.
answered Jan 8 at 12:55
DavidGDavidG
2,1671720
$endgroup$
add a comment |
$begingroup$
NOT A SOLUTION:
Not sure if this will be of help. I would start by repositioning the integrand:
begin{equation}
I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
end{equation}
Now let $t = dfrac{x}{4 - x}$ :
begin{equation}
I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
end{equation}
I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.
begin{equation}
lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
end{equation}
Then,
begin{align}
I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
end{align}
But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.
answered Jan 8 at 12:55
DavidGDavidG
2,1671720
$endgroup$
add a comment |
$begingroup$
NOT A SOLUTION:
Not sure if this will be of help. I would start by repositioning the integrand:
begin{equation}
I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
end{equation}
Now let $t = dfrac{x}{4 - x}$ :
begin{equation}
I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
end{equation}
I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.
begin{equation}
lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
end{equation}
Then,
begin{align}
I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
end{align}
But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.
answered Jan 8 at 12:55
DavidGDavidG
2,1671720
$endgroup$
NOT A SOLUTION:
Not sure if this will be of help. I would start by repositioning the integrand:
begin{equation}
I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
end{equation}
Now let $t = dfrac{x}{4 - x}$ :
begin{equation}
I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
end{equation}
I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.
begin{equation}
lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
end{equation}
Then,
begin{align}
I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
end{align}
But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.
answered Jan 8 at 12:55
DavidGDavidG
2,1671720
edited Jan 8 at 13:01
answered Jan 8 at 12:55
DavidGDavidG
2,1671720
answered Jan 8 at 12:55
DavidGDavidG
2,1671720
answered Jan 8 at 12:55
DavidGDavidG
2,1671720
2,1671720
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$begingroup$
I guess it should be $ntoinfty$, not $x$.
$endgroup$
– Mark
Jan 7 at 14:34
1
$begingroup$
Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
$endgroup$
– Math_QED
Jan 7 at 14:56
$begingroup$
@Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
$endgroup$
– Mark
Jan 7 at 15:18