limit of x sin( 1/x ) as x approaches infinity
Replace with u = 1/x, x = 1/u
As x approaches infinity, u approaches zero
limit of x sin( 1/x ) as x approaches infinity =
limit of ( 1/u )( sin( u ) ) as u approaches zero
By definition,
http://en.wikipedia.org/wiki/Sine#Sine
sin( u ) = u - ( u ^ 3 ) / 3! + ( u ^ 5 ) / 5! - ( u ^ 7 ) / 7! + ...
( 1/u )( sin( u ) ) =
( 1/u )( u - ( u ^ 3 ) / 3! + ( u ^ 5 ) / 5! - ( u ^ 7 ) / 7! + ... ) =
1 - ( u ^ 2 ) / 3! + ( u ^ 4 ) / 5! - ( u ^ 6 ) / 7! + ...
limit of x sin( 1/x ) as x approaches infinity =
limit of ( 1/u )( sin( u ) ) as u approaches zero =
limit of ( 1 - ( u ^ 2 ) / 3! + ( u ^ 4 ) / 5! - ( u ^ 6 ) / 7! + ... ) as u approaches zero =
1 - ( 0 ^ 2 ) / 3! + ( 0 ^ 4 ) / 5! - ( 0 ^ 6 ) / 7 ! + ... =
1 - 0 + 0 - 0 + ... =
1
