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Question-78635

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Question Number 78635 by Pratah last updated on 19/Jan/20
Commented by john santu last updated on 19/Jan/20
lim_(n→∞) (((∫_n ^(2n) ((x/(x^5 +1)))dx )/n^(−3) ))=lim_(n→∞) (((((2n)/(32n^5 +1))).2−((n/(n^5 +1))).1)/(−3n^(−4) )) =−(1/3)lim_(n→∞) (((4n^5 )/(32n^5 +1)))−((n^5 /(n^5 +1))) = −(1/3)((1/8)−1)= −(1/3)×(((−7)/8))=(7/(24)).
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\underset{{n}} {\overset{\mathrm{2}{n}} {\int}}\:\left(\frac{{x}}{{x}^{\mathrm{5}} +\mathrm{1}}\right){dx}\:}{{n}^{−\mathrm{3}} }\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{2}{n}}{\mathrm{32}{n}^{\mathrm{5}} +\mathrm{1}}\right).\mathrm{2}−\left(\frac{{n}}{{n}^{\mathrm{5}} +\mathrm{1}}\right).\mathrm{1}}{−\mathrm{3}{n}^{−\mathrm{4}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{4}{n}^{\mathrm{5}} }{\mathrm{32}{n}^{\mathrm{5}} +\mathrm{1}}\right)−\left(\frac{{n}^{\mathrm{5}} }{{n}^{\mathrm{5}} +\mathrm{1}}\right) \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{1}\right)=\:−\frac{\mathrm{1}}{\mathrm{3}}×\left(\frac{−\mathrm{7}}{\mathrm{8}}\right)=\frac{\mathrm{7}}{\mathrm{24}}. \\ $$
Commented by Pratah last updated on 19/Jan/20
thanks
$$\mathrm{thanks} \\ $$
Commented by mr W last updated on 19/Jan/20
it is to prove at first thatlim_(n→∞) ∫_n ^(2n) ((x/(x^5 +1)))dx=0!
$${it}\:{is}\:{to}\:{prove}\:{at}\:{first}\:{that}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{n}} {\overset{\mathrm{2}{n}} {\int}}\:\left(\frac{{x}}{{x}^{\mathrm{5}} +\mathrm{1}}\right){dx}=\mathrm{0}! \\ $$
Commented by jagoll last updated on 19/Jan/20
L′Hopital can be using for limit form (∞/∞)
$$\mathrm{L}'\mathrm{Hopital}\:\mathrm{can}\:\mathrm{be}\:\mathrm{using}\: \\ $$$$\mathrm{for}\:\mathrm{limit}\:\mathrm{form}\:\frac{\infty}{\infty} \\ $$
Commented by mr W last updated on 19/Jan/20
we have here (0/0)
$${we}\:{have}\:{here}\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$
Commented by john santu last updated on 19/Jan/20
in other way let n = (1/t) ⇒t →0 lim_(t→0) (((∫^(2/t) _(1/t) ((x/(x^5 +1)))dx)/t^3 ))= lim_(t→0) [((((2/t)).(((−2)/t^2 )))/((((32)/t^5 )+1)))−((((1/t)).(((−1)/t^2 )))/(((1/t^5 )+1)))] ×(1/(3t^2 )) = [((−4t^2 )/(32+t^5 ))−(((−t^2 ))/((1+t^5 )))]×(1/(3t^2 )) = [((−1)/8)+1]×(1/3)=(7/(24)) .★
$${in}\:{other}\:{way}\: \\ $$$${let}\:{n}\:=\:\frac{\mathrm{1}}{{t}}\:\Rightarrow{t}\:\rightarrow\mathrm{0} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\underset{\frac{\mathrm{1}}{{t}}} {\int}^{\frac{\mathrm{2}}{{t}}} \left(\frac{{x}}{{x}^{\mathrm{5}} +\mathrm{1}}\right){dx}}{{t}^{\mathrm{3}} }\right)= \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\left(\frac{\mathrm{2}}{{t}}\right).\left(\frac{−\mathrm{2}}{{t}^{\mathrm{2}} }\right)}{\left(\frac{\mathrm{32}}{{t}^{\mathrm{5}} }+\mathrm{1}\right)}−\frac{\left(\frac{\mathrm{1}}{{t}}\right).\left(\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{\left(\frac{\mathrm{1}}{{t}^{\mathrm{5}} }+\mathrm{1}\right)}\right]\:×\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$=\:\left[\frac{−\mathrm{4}{t}^{\mathrm{2}} }{\mathrm{32}+{t}^{\mathrm{5}} }−\frac{\left(−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{5}} \right)}\right]×\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$=\:\left[\frac{−\mathrm{1}}{\mathrm{8}}+\mathrm{1}\right]×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{7}}{\mathrm{24}}\:.\bigstar \\ $$
Commented by mr W last updated on 19/Jan/20
I_n =∫_n ^(2n) ((x/(x^5 +1)))dx 0<(x/(x^5 +1))<(x/x^5 )=(1/x^4 ) 0<I_n <(1/((2n)^4 ))=(1/(32n^4 )) 0<lim_(n→∞) I_n <lim_(n→∞) (1/(32n^4 ))=0 ⇒lim_(n→∞) I_n =0 ⇒lim_(n→∞) ((∫_n ^(2n) ((x/(x^5 +1)))dx)/(1/n^3 ))=(0/0)
$${I}_{{n}} =\underset{{n}} {\overset{\mathrm{2}{n}} {\int}}\:\left(\frac{{x}}{{x}^{\mathrm{5}} +\mathrm{1}}\right){dx} \\ $$$$\mathrm{0}<\frac{{x}}{{x}^{\mathrm{5}} +\mathrm{1}}<\frac{{x}}{{x}^{\mathrm{5}} }=\frac{\mathrm{1}}{{x}^{\mathrm{4}} } \\ $$$$\mathrm{0}<{I}_{{n}} <\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{32}{n}^{\mathrm{4}} } \\ $$$$\mathrm{0}<\underset{{n}\rightarrow\infty} {\mathrm{lim}}{I}_{{n}} <\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{32}{n}^{\mathrm{4}} }=\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{I}_{{n}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\underset{{n}} {\overset{\mathrm{2}{n}} {\int}}\:\left(\frac{{x}}{{x}^{\mathrm{5}} +\mathrm{1}}\right){dx}}{\frac{\mathrm{1}}{{n}^{\mathrm{3}} }}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$
Commented by john santu last updated on 19/Jan/20
by using the sequence theorem sir
$${by}\:{using}\:{the}\:{sequence}\:{theorem}\:{sir} \\ $$
Answered by mr W last updated on 19/Jan/20
I=∫_n ^(2n) (x/(1+x^5 ))dx let t=(1/x) dx=−(dt/t^2 ) I=∫_(1/n) ^(1/(2n)) ((1/t)/(1+(1/t^5 )))×(−(1/t^2 ))dt =∫_(1/(2n)) ^(1/n) (t^2 /(1+t^5 ))dt lim_(n→∞) n^3 ∫_n ^(2n) (x/(1+x^5 ))dx =lim_(n→∞) ((∫_n ^(2n) (x/(1+x^5 ))dx)/(((1/n))^3 )) =lim_(n→∞) ((∫_(1/(2n)) ^(1/n) (t^2 /(1+t^5 ))dt)/(((1/n))^3 )) =lim_(u→0) ((∫_(u/2) ^u (t^2 /(1+t^5 ))dt)/u^3 ) (=(0/0)) =lim_(u→0) (((u^2 /(1+u^5 ))−((((u/2))^2 )/(1+((u/2))^5 ))×(1/2))/(3u^2 )) =lim_(u→0) ((((1/(1+u^5 ))−(4/(32+u^5 ))))/3) =(7/(8×3)) =(7/(24))
$${I}=\int_{{n}} ^{\mathrm{2}{n}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{5}} }{dx} \\ $$$${let}\:{t}=\frac{\mathrm{1}}{{x}} \\ $$$${dx}=−\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$${I}=\int_{\frac{\mathrm{1}}{{n}}} ^{\frac{\mathrm{1}}{\mathrm{2}{n}}} \frac{\frac{\mathrm{1}}{{t}}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{5}} }}×\left(−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}{n}}} ^{\frac{\mathrm{1}}{{n}}} \frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{5}} }{dt} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}^{\mathrm{3}} \int_{{n}} ^{\mathrm{2}{n}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{5}} }{dx} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\int_{{n}} ^{\mathrm{2}{n}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{5}} }{dx}}{\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{3}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\int_{\frac{\mathrm{1}}{\mathrm{2}{n}}} ^{\frac{\mathrm{1}}{{n}}} \frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{5}} }{dt}}{\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{3}} } \\ $$$$=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{\frac{{u}}{\mathrm{2}}} ^{{u}} \frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{5}} }{dt}}{{u}^{\mathrm{3}} }\:\:\:\:\:\left(=\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{5}} }−\frac{\left(\frac{{u}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{u}}{\mathrm{2}}\right)^{\mathrm{5}} }×\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{3}{u}^{\mathrm{2}} } \\ $$$$=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{5}} }−\frac{\mathrm{4}}{\mathrm{32}+{u}^{\mathrm{5}} }\right)}{\mathrm{3}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{8}×\mathrm{3}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{24}} \\ $$

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