Find limit in use of integralshow to find the limit?Find limit $lim_{x rightarrow 0} ,...
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Find limit in use of integrals
how to find the limit?Find limit $lim_{x rightarrow 0} , frac{arctan(3x)}{tanbig((x+3pi)/3big)}$ without using L'Hospital's rulemathematical analysisLimit of a complex sequenceUsing the mean value theorem calculate this limit : $lim_{x rightarrow 0} frac{arctan(x^2+x-1)+frac{pi}{4}}{x^2+3x}$Limit of a definite integral with parameter (2)Limit of $(intlimits_0^n (1+arctan^2x ),dx )^ {frac{1}{n}}$limit of and integral depending on nA tricky limit involving exponential integralsFind limit supremum from 3 sequence theorem
$begingroup$
Find limit $lim_{n rightarrow infty} int_{-1}^{1} x^5 cdot arctan{(nx)} dx $
From mean-value-theorem we have
$$ frac{1}{2} c^5 cdot arctan{(nc)} mbox{ for some c } in (-1,1) $$
$$ underbrace{frac{1}{2} cdot arctan{(-n)}}_{rightarrow - pi /4} le frac{1}{2} c^5 cdot arctan{(nc)} le underbrace{frac{1}{2} cdot arctan{(n)}}_{rightarrow pi /4} $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
integration limits
asked 8 hours ago
Trebacz112Trebacz112
405
New contributor
Trebacz112 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Find limit $lim_{n rightarrow infty} int_{-1}^{1} x^5 cdot arctan{(nx)} dx $
From mean-value-theorem we have
$$ frac{1}{2} c^5 cdot arctan{(nc)} mbox{ for some c } in (-1,1) $$
$$ underbrace{frac{1}{2} cdot arctan{(-n)}}_{rightarrow - pi /4} le frac{1}{2} c^5 cdot arctan{(nc)} le underbrace{frac{1}{2} cdot arctan{(n)}}_{rightarrow pi /4} $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
integration limits
asked 8 hours ago
Trebacz112Trebacz112
405
New contributor
Trebacz112 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
8 hours ago
add a comment |
$begingroup$
Find limit $lim_{n rightarrow infty} int_{-1}^{1} x^5 cdot arctan{(nx)} dx $
From mean-value-theorem we have
$$ frac{1}{2} c^5 cdot arctan{(nc)} mbox{ for some c } in (-1,1) $$
$$ underbrace{frac{1}{2} cdot arctan{(-n)}}_{rightarrow - pi /4} le frac{1}{2} c^5 cdot arctan{(nc)} le underbrace{frac{1}{2} cdot arctan{(n)}}_{rightarrow pi /4} $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
integration limits
asked 8 hours ago
Trebacz112Trebacz112
405
New contributor
Trebacz112 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Find limit $lim_{n rightarrow infty} int_{-1}^{1} x^5 cdot arctan{(nx)} dx $
From mean-value-theorem we have
$$ frac{1}{2} c^5 cdot arctan{(nc)} mbox{ for some c } in (-1,1) $$
$$ underbrace{frac{1}{2} cdot arctan{(-n)}}_{rightarrow - pi /4} le frac{1}{2} c^5 cdot arctan{(nc)} le underbrace{frac{1}{2} cdot arctan{(n)}}_{rightarrow pi /4} $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
integration limits
integration limits
asked 8 hours ago
Trebacz112Trebacz112
405
New contributor
Trebacz112 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Trebacz112Trebacz112
405
New contributor
Trebacz112 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Trebacz112Trebacz112
405
New contributor
Trebacz112 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Trebacz112Trebacz112
405
asked 8 hours ago
Trebacz112Trebacz112
405
405
New contributor
Trebacz112 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Trebacz112 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
8 hours ago
add a comment |
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
8 hours ago
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
8 hours ago
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=frac{y^{6}}{6}arctan y-frac{1}{6}intfrac{y^{6}}{1+y^{2}}dy\=frac{y^{6}}{6}arctan y-frac{1}{6}intleft(y^{4}-y^{2}+1-frac{1}{1+y^{2}}right)dy\=frac{y^6+1}{6}arctan y-frac{1}{30}y^5+frac{1}{18}y^3-frac16 y+C.$$Hence $$frac{1}{n^6}int_{-n}^n y^5arctan ydy=frac{frac{n^6}{3}arctan n+o(n^6)}{n^6}stackrel{ntoinfty}{to}frac{1}{3}arctaninfty=frac{pi}{6}.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_{ntoinfty}int_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=frac{pi}{6}.$$
answered 8 hours ago
J.G.J.G.
38.4k23757
$endgroup$
add a comment |
$begingroup$
Hint: For the integral $$int_{-1}^{1}x^5arctan(nx)dx$$ we get
$$frac{15 left(n^6+1right) tan ^{-1}(n)-3 n^5+5
n^3-15 n}{45 n^6}$$
answered 8 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.5k42867
$endgroup$
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
8 hours ago
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
8 hours ago
1
$begingroup$
Hint: $$frac{x^6}{n^2x^2+1}={frac {{x}^{4}}{{n}^{2}}}-{frac {{x}^{2}}{{n}^{4}}}+{n}^{-6}-{frac {1}{{n}^{6} left( {n}^{2}{x}^{2}+1 right) }} $$
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
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votes
$begingroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=frac{y^{6}}{6}arctan y-frac{1}{6}intfrac{y^{6}}{1+y^{2}}dy\=frac{y^{6}}{6}arctan y-frac{1}{6}intleft(y^{4}-y^{2}+1-frac{1}{1+y^{2}}right)dy\=frac{y^6+1}{6}arctan y-frac{1}{30}y^5+frac{1}{18}y^3-frac16 y+C.$$Hence $$frac{1}{n^6}int_{-n}^n y^5arctan ydy=frac{frac{n^6}{3}arctan n+o(n^6)}{n^6}stackrel{ntoinfty}{to}frac{1}{3}arctaninfty=frac{pi}{6}.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_{ntoinfty}int_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=frac{pi}{6}.$$
answered 8 hours ago
J.G.J.G.
38.4k23757
$endgroup$
add a comment |
$begingroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=frac{y^{6}}{6}arctan y-frac{1}{6}intfrac{y^{6}}{1+y^{2}}dy\=frac{y^{6}}{6}arctan y-frac{1}{6}intleft(y^{4}-y^{2}+1-frac{1}{1+y^{2}}right)dy\=frac{y^6+1}{6}arctan y-frac{1}{30}y^5+frac{1}{18}y^3-frac16 y+C.$$Hence $$frac{1}{n^6}int_{-n}^n y^5arctan ydy=frac{frac{n^6}{3}arctan n+o(n^6)}{n^6}stackrel{ntoinfty}{to}frac{1}{3}arctaninfty=frac{pi}{6}.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_{ntoinfty}int_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=frac{pi}{6}.$$
answered 8 hours ago
J.G.J.G.
38.4k23757
$endgroup$
add a comment |
$begingroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=frac{y^{6}}{6}arctan y-frac{1}{6}intfrac{y^{6}}{1+y^{2}}dy\=frac{y^{6}}{6}arctan y-frac{1}{6}intleft(y^{4}-y^{2}+1-frac{1}{1+y^{2}}right)dy\=frac{y^6+1}{6}arctan y-frac{1}{30}y^5+frac{1}{18}y^3-frac16 y+C.$$Hence $$frac{1}{n^6}int_{-n}^n y^5arctan ydy=frac{frac{n^6}{3}arctan n+o(n^6)}{n^6}stackrel{ntoinfty}{to}frac{1}{3}arctaninfty=frac{pi}{6}.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_{ntoinfty}int_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=frac{pi}{6}.$$
answered 8 hours ago
J.G.J.G.
38.4k23757
$endgroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=frac{y^{6}}{6}arctan y-frac{1}{6}intfrac{y^{6}}{1+y^{2}}dy\=frac{y^{6}}{6}arctan y-frac{1}{6}intleft(y^{4}-y^{2}+1-frac{1}{1+y^{2}}right)dy\=frac{y^6+1}{6}arctan y-frac{1}{30}y^5+frac{1}{18}y^3-frac16 y+C.$$Hence $$frac{1}{n^6}int_{-n}^n y^5arctan ydy=frac{frac{n^6}{3}arctan n+o(n^6)}{n^6}stackrel{ntoinfty}{to}frac{1}{3}arctaninfty=frac{pi}{6}.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_{ntoinfty}int_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=frac{pi}{6}.$$
answered 8 hours ago
J.G.J.G.
38.4k23757
edited 8 hours ago
answered 8 hours ago
J.G.J.G.
38.4k23757
answered 8 hours ago
J.G.J.G.
38.4k23757
answered 8 hours ago
J.G.J.G.
38.4k23757
38.4k23757
add a comment |
add a comment |
$begingroup$
Hint: For the integral $$int_{-1}^{1}x^5arctan(nx)dx$$ we get
$$frac{15 left(n^6+1right) tan ^{-1}(n)-3 n^5+5
n^3-15 n}{45 n^6}$$
answered 8 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.5k42867
$endgroup$
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
8 hours ago
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
8 hours ago
1
$begingroup$
Hint: $$frac{x^6}{n^2x^2+1}={frac {{x}^{4}}{{n}^{2}}}-{frac {{x}^{2}}{{n}^{4}}}+{n}^{-6}-{frac {1}{{n}^{6} left( {n}^{2}{x}^{2}+1 right) }} $$
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
add a comment |
$begingroup$
Hint: For the integral $$int_{-1}^{1}x^5arctan(nx)dx$$ we get
$$frac{15 left(n^6+1right) tan ^{-1}(n)-3 n^5+5
n^3-15 n}{45 n^6}$$
answered 8 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.5k42867
$endgroup$
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
8 hours ago
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
8 hours ago
1
$begingroup$
Hint: $$frac{x^6}{n^2x^2+1}={frac {{x}^{4}}{{n}^{2}}}-{frac {{x}^{2}}{{n}^{4}}}+{n}^{-6}-{frac {1}{{n}^{6} left( {n}^{2}{x}^{2}+1 right) }} $$
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
add a comment |
$begingroup$
Hint: For the integral $$int_{-1}^{1}x^5arctan(nx)dx$$ we get
$$frac{15 left(n^6+1right) tan ^{-1}(n)-3 n^5+5
n^3-15 n}{45 n^6}$$
answered 8 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.5k42867
$endgroup$
Hint: For the integral $$int_{-1}^{1}x^5arctan(nx)dx$$ we get
$$frac{15 left(n^6+1right) tan ^{-1}(n)-3 n^5+5
n^3-15 n}{45 n^6}$$
answered 8 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.5k42867
answered 8 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.5k42867
answered 8 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.5k42867
answered 8 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.5k42867
81.5k42867
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
8 hours ago
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
8 hours ago
1
$begingroup$
Hint: $$frac{x^6}{n^2x^2+1}={frac {{x}^{4}}{{n}^{2}}}-{frac {{x}^{2}}{{n}^{4}}}+{n}^{-6}-{frac {1}{{n}^{6} left( {n}^{2}{x}^{2}+1 right) }} $$
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
add a comment |
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
8 hours ago
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
8 hours ago
1
$begingroup$
Hint: $$frac{x^6}{n^2x^2+1}={frac {{x}^{4}}{{n}^{2}}}-{frac {{x}^{2}}{{n}^{4}}}+{n}^{-6}-{frac {1}{{n}^{6} left( {n}^{2}{x}^{2}+1 right) }} $$
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
8 hours ago
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
8 hours ago
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
8 hours ago
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
8 hours ago
1
1
$begingroup$
Hint: $$frac{x^6}{n^2x^2+1}={frac {{x}^{4}}{{n}^{2}}}-{frac {{x}^{2}}{{n}^{4}}}+{n}^{-6}-{frac {1}{{n}^{6} left( {n}^{2}{x}^{2}+1 right) }} $$
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
Hint: $$frac{x^6}{n^2x^2+1}={frac {{x}^{4}}{{n}^{2}}}-{frac {{x}^{2}}{{n}^{4}}}+{n}^{-6}-{frac {1}{{n}^{6} left( {n}^{2}{x}^{2}+1 right) }} $$
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
add a comment |
Trebacz112 is a new contributor. Be nice, and check out our Code of Conduct.
Trebacz112 is a new contributor. Be nice, and check out our Code of Conduct.
Trebacz112 is a new contributor. Be nice, and check out our Code of Conduct.
Trebacz112 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
8 hours ago