Exercise 3.  State why   [Graphics:Images/ComplexFunLimitModHome_gr_158.gif].  

Solution 3.

See text and/or instructor's solution manual.

Solution.  For the real functions  [Graphics:../Images/ComplexFunLimitModHome_gr_159.gif]  and  [Graphics:../Images/ComplexFunLimitModHome_gr_160.gif]  we have:

        [Graphics:../Images/ComplexFunLimitModHome_gr_161.gif][Graphics:../Images/ComplexFunLimitModHome_gr_162.gif],  
    and
        [Graphics:../Images/ComplexFunLimitModHome_gr_163.gif][Graphics:../Images/ComplexFunLimitModHome_gr_164.gif].  

The result now follows by Theorem 2.1 since the real and imaginary parts of the last expression have limits that imply the desired conclusion.

        [Graphics:../Images/ComplexFunLimitModHome_gr_165.gif]  

Therefore  [Graphics:../Images/ComplexFunLimitModHome_gr_166.gif]  is continuous for all  [Graphics:../Images/ComplexFunLimitModHome_gr_167.gif].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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