Limit with integral

Calculus Level pending

lim x 0 ( 1 + 0 a x 1 ( sin 2 arctan t ) ( 1 + t 2 ) ln a d t ) 1 x = ? \large \lim _{ x\to 0 }\left(1+\int _0 ^{ \sqrt { a^x-1 }}(\sin 2\arctan t) (1+t^2)^{ \ln a } \, dt \right)^{ \frac 1x} =\, ?

where a > 1 a>1 and x > 0 x>0 .

e a { e }^{ a } 1 1 a a ln a \ln { a } e e

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1 solution

Patrik Kovacs
Sep 30, 2016

L = e lim x 0 0 a x 1 ( sin 2 arctan t ) ( 1 + t 2 ) ln a d t x = e l sin 2 arctan t = 2 t 1 + t 2 = [ ln ( 1 + t 2 ) ] ( 1 + t 2 ) ln a = a ln ( 1 + t 2 ) 0 a x 1 ( sin 2 arctan t ) ( 1 + t 2 ) ln a d t = a x ln a 1 ln a l = lim x 0 a x ln a 1 x ln a = ln a L = e ln a = a L={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \int _{ 0 }^{ \sqrt { { a }^{ x }-1 } }{ (\sin { 2\arctan { t) } { (1+{ t }^{ 2 }) }^{ \ln { a } }dt } } }{ x } } }={ e }^{ l }\\ \sin { 2\arctan { t } =\quad \frac { 2t }{ 1+{ t }^{ 2 } } } =\quad [\ln { (1+{ t }^{ 2 })] } '\\ { (1+{ t }^{ 2 }) }^{ \ln { a } }\quad =\quad { a }^{ \ln { (1+{ t }^{ 2 }) } }\\ \int _{ 0 }^{ \sqrt { { a }^{ x }-1 } }{ (\sin { 2\arctan { t) } { (1+{ t }^{ 2 }) }^{ \ln { a } }dt } } =\frac { { a }^{ x\ln { a } }-1 }{ \ln { a } } \\ l=\lim _{ x\rightarrow 0 }{ \frac { { a }^{ x\ln { a } }-1 }{ x\ln { a } } } =\ln { a } \\ L={ e }^{ \ln { a } }=a\\

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