$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$ [duplicate]
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Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$
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Suppose the limit $$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$$
and is larger than $frac{1}{2}$. Then
(A) $0.5<L<2$
(B) $2<L<3$
(C) $3<L<4$
(D) $Lge 4$
I used $x=tantheta$ but not getting the value
limits
marked as duplicate by Nosrati, Mark Viola
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Nov 14 at 18:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$
3 answers
Suppose the limit $$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$$
and is larger than $frac{1}{2}$. Then
(A) $0.5<L<2$
(B) $2<L<3$
(C) $3<L<4$
(D) $Lge 4$
I used $x=tantheta$ but not getting the value
limits
marked as duplicate by Nosrati, Mark Viola
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Nov 14 at 18:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
– StubbornAtom
Nov 14 at 17:20
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up vote
1
down vote
favorite
This question already has an answer here:
Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$
3 answers
Suppose the limit $$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$$
and is larger than $frac{1}{2}$. Then
(A) $0.5<L<2$
(B) $2<L<3$
(C) $3<L<4$
(D) $Lge 4$
I used $x=tantheta$ but not getting the value
limits
This question already has an answer here:
Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$
3 answers
Suppose the limit $$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$$
and is larger than $frac{1}{2}$. Then
(A) $0.5<L<2$
(B) $2<L<3$
(C) $3<L<4$
(D) $Lge 4$
I used $x=tantheta$ but not getting the value
This question already has an answer here:
Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$
3 answers
limits
limits
edited Nov 14 at 17:18
Tianlalu
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asked Nov 14 at 17:16
Samar Imam Zaidi
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1,427417
marked as duplicate by Nosrati, Mark Viola
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Nov 14 at 18:52
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Nov 14 at 18:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
– StubbornAtom
Nov 14 at 17:20
add a comment |
1
For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
– StubbornAtom
Nov 14 at 17:20
1
1
For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
– StubbornAtom
Nov 14 at 17:20
For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
– StubbornAtom
Nov 14 at 17:20
add a comment |
2 Answers
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For $x ge 0$ you have
$$(1+x^2)^n ge 1+nx^2$$ hence
$$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and
$$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
$$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$
Therefore if $L$ exists as supposed in the hypothesis, we have
$$L le frac{pi}{2} < 2$$
proving that the right answer is (A).
Nice, that's simpler than my answer. +1
– zhw.
Nov 14 at 19:59
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For $xin [0,1],$
$$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$
Thus our expression is bounded above by
$$sqrt nint_0^1 (1-x^2/2)^n, dx.$$
Let $x=y/sqrt n.$ The above turns into
$$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$
add a comment |
2 Answers
2
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2 Answers
2
active
oldest
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active
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active
oldest
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up vote
2
down vote
For $x ge 0$ you have
$$(1+x^2)^n ge 1+nx^2$$ hence
$$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and
$$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
$$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$
Therefore if $L$ exists as supposed in the hypothesis, we have
$$L le frac{pi}{2} < 2$$
proving that the right answer is (A).
Nice, that's simpler than my answer. +1
– zhw.
Nov 14 at 19:59
add a comment |
up vote
2
down vote
For $x ge 0$ you have
$$(1+x^2)^n ge 1+nx^2$$ hence
$$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and
$$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
$$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$
Therefore if $L$ exists as supposed in the hypothesis, we have
$$L le frac{pi}{2} < 2$$
proving that the right answer is (A).
Nice, that's simpler than my answer. +1
– zhw.
Nov 14 at 19:59
add a comment |
up vote
2
down vote
up vote
2
down vote
For $x ge 0$ you have
$$(1+x^2)^n ge 1+nx^2$$ hence
$$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and
$$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
$$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$
Therefore if $L$ exists as supposed in the hypothesis, we have
$$L le frac{pi}{2} < 2$$
proving that the right answer is (A).
For $x ge 0$ you have
$$(1+x^2)^n ge 1+nx^2$$ hence
$$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and
$$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
$$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$
Therefore if $L$ exists as supposed in the hypothesis, we have
$$L le frac{pi}{2} < 2$$
proving that the right answer is (A).
edited Nov 14 at 18:11
answered Nov 14 at 18:05
mathcounterexamples.net
23.6k21753
23.6k21753
Nice, that's simpler than my answer. +1
– zhw.
Nov 14 at 19:59
add a comment |
Nice, that's simpler than my answer. +1
– zhw.
Nov 14 at 19:59
Nice, that's simpler than my answer. +1
– zhw.
Nov 14 at 19:59
Nice, that's simpler than my answer. +1
– zhw.
Nov 14 at 19:59
add a comment |
up vote
0
down vote
For $xin [0,1],$
$$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$
Thus our expression is bounded above by
$$sqrt nint_0^1 (1-x^2/2)^n, dx.$$
Let $x=y/sqrt n.$ The above turns into
$$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$
add a comment |
up vote
0
down vote
For $xin [0,1],$
$$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$
Thus our expression is bounded above by
$$sqrt nint_0^1 (1-x^2/2)^n, dx.$$
Let $x=y/sqrt n.$ The above turns into
$$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$
add a comment |
up vote
0
down vote
up vote
0
down vote
For $xin [0,1],$
$$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$
Thus our expression is bounded above by
$$sqrt nint_0^1 (1-x^2/2)^n, dx.$$
Let $x=y/sqrt n.$ The above turns into
$$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$
For $xin [0,1],$
$$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$
Thus our expression is bounded above by
$$sqrt nint_0^1 (1-x^2/2)^n, dx.$$
Let $x=y/sqrt n.$ The above turns into
$$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$
answered Nov 14 at 18:21
zhw.
70.4k42975
70.4k42975
add a comment |
add a comment |
1
For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
– StubbornAtom
Nov 14 at 17:20