what is the limit of xsin(1/x) as x approaches infinity?

[mred_11]

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

 

[nouras]

please show steps and dont use L'Hopitals rule!

 

[TedS]

answer is " 1 ".... let w = 1 / x, w ---> 0 as x --> ?.....problem is now sin w / w---> 1 as w ---> 0 , familiar result in calculus