Show that 0sin1x does not exist lim x SolutionTaking 1 2 in

Solution

Taking = 1 2 in the definition of this limit we find that there is a > 0 such that |sin( 1 x )L| < 1 2 whenever 0 < x < . Now find an integer n > 0 so large that x1 = 1/(2n/2) < . Then also x2 = 1/(2n+/2) < x1 < . Thus we can conclude that both |sin( 1 x1 )L| and |sin( 1 x2 )L| are less than 1 2 . But x1 and x2 were chosen so that sin( 1 x1 ) = 1 and sin( 1 x2 ) = 1. Thus we have | 1L| < 1 2 and |1L| < 1 2 , in other words both 1 and 1 lie in the interval (L 1 2 ,L+ 1 2 ). Since this interval has length 1 we conclude that the distance from 1 to 1 is less than 1, which is certainly not true. This contradiction shows that our assumption that the limit existed was false, that is the limit does not exist. Remark: Why did we choose = 1 2 ? Actually any 1 would work for us here since it would lead to the contradiction that the distance from 1 to 1 is less than 2. This is a contradiction but only barely a contradiction, so we used = 1 2 to make the contradiction starker, thus (hopefully) improving your intuition about this argument. Theoretically, however = 1 2 is no better than = 1, they both do the job. A careful inspection of the proof of Example 11 reveals that we used the following fact: if R is a positive real number then there is a positive integer n such that n > R. (In order to make 1/(2n/2) < we must make n > ( 1+ 2 ) 2 = R.) This is called the Archimedean property of the real numbers.