Limitless?

Calculus Level 3

lim x 0 ( sin x + cos x ) 1 x = ? \large \lim_{ x \to 0 } (\sin x +\cos x)^ \frac 1x = \, ?

π 2 { \pi }^{ 2 } e 2 { e }^{ 2 } π \pi e e

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1 solution

Mohammad Hamdar
Feb 6, 2017

sin x + cos x = 1 + sin x + cos x 1 \sin { x } +\cos { x } =1+\sin { x } +\cos { x } -1 = 1 + 2 sin x 2 cos x 2 2 sin 2 x 2 1+2\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } } -2\sin ^{ 2 }{ \frac { x }{ 2 } } } = 1 + h ( x ) sin x 2 1+h\left( x \right) \sin { \frac { x }{ 2 } } with L i m x 0 h ( x ) = L i m x 0 2 ( cos x 2 sin x 2 ) = 2 { Lim }_{ x\rightarrow 0 }h\left( x \right) ={ Lim }_{ x\rightarrow 0 }2(\cos { \frac { x }{ 2 } -\sin { \frac { x }{ 2 } )=2 } } . Then, L i m x 0 1 x ln ( sin x + cos x ) { Lim }_{ x\rightarrow 0 }\frac { 1 }{ x } \ln { (\sin { x } } +\cos { x } ) = L i m x 0 [ ln ( 1 + h ( x ) sin x 2 ) h ( x ) sin x 2 sin x 2 x 2 h ( x ) 2 ] { Lim }_{ x\rightarrow 0 }[\frac { \ln { (1+h\left( x \right) \sin { \frac { x }{ 2 } ) } } }{ h\left( x \right) \sin { \frac { x }{ 2 } } } \frac { \sin { \frac { x }{ 2 } } }{ \frac { x }{ 2 } } \frac { h\left( x \right) }{ 2 } ] = 1 =1 Hence, L i m x 0 ( sin x + cos x ) 1 x = e { Lim }_{ x\rightarrow 0 }{ (\sin { x } +\cos { x) } }^{ \frac { 1 }{ x } }=e

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