Solution 1 (a).

Answer.   No.  

Solution.   No.  Otherwise, since     that is analytic at    we have the limits

                    ,    and  

                    .  

This implies that    which is impossible.

Therefore, there does not exist a function    that is analytic at    such that    and  .

Alternate Solution.   No.  Since is analytic at , there exists a neighborhood    in which is analytic,

and we can choose an integer    so that    and  .  

Apply Corollary 7.10 for the sequences  ,  respectively.

For the sequence   ,   and the function      we have

                    ,   

and Corollary 7.10 implies that   .  

For the sequence   ,   and the function      we have

                    ,   

and Corollary 7.10 implies that   .  

Since   ,   this is impossible.

Therefore, there does not exist a function    that is analytic at    such that    and  .

This solution is complements of the authors.

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