Exercise 15.  Prove that a sequence can have only one limit.  

Hint.  Suppose that there is a sequence    such that      and   .   

Show this implies      by proving that for all   ,   we must have   .  

Solution 15.

See text and/or instructor's solution manual.

Solution.   Following the hint, for    there exists numbers    such that


                       implies that   ,   and  

                       implies that   .    

Let   .   Then      implies that  

                      

Since       is arbitrary, this implies that .  

Therefore   .  

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell