2.3  Limits and Continuity

    Let  u = u(x,y)  be a real-valued function of the two real variables x and y.  Recall that u has the limit as approaches provided that the value of can be made to get as close as we please to the value by taking to be sufficiently close to .  When this happens we write  

            .  

    In more technical language, u has the limit as (x,y) approaches    iff    can be made arbitrarily small by making both    and    small. This condition is like the definition of a limit for functions of one variable. The point is in the xy plane, and the distance between and is .   With this perspective we can now give a precise definition of a limit.

Definition 2.3 ( limit of u(x,y) ).  The expression    means that for each number , there corresponds a number such that  

(2-15)               whenever   .  

Example 2.14.  Show, if  ,  then  .  

Solution.  If  ,  and    then  

            .  

Because    and because  ,  we have

              whenever  .

Hence, for any  ,  Inequality (2-15) is satisfied for  ;  that is, has the limit    as approaches .

Explore Solution 2.14.

    The value of the limit must not depend on how (x,y) approaches , so must approach the value when approaches along any curve that ends at the point .  Conversely, if we can find two curves that end at along which approaches the two distinct values , respectively, then does not have a limit as approaches .  

Example 2.15.  Show that the function    does not have a limit as (x,y) approaches .

Solution.  If we let (x,y) approach (0,0) along the x axis, then  

            .  

But if we let (x,y) approach (0,0) along the line , then  

            .  

Because the value of the limit differs depending on how (x,y) approaches  (0,0), we conclude that does not have a limit as approaches .

Explore Solution 2.15.

    Let f(z) be a complex function of the complex variable z that is defined for all values of z in some neighborhood of , except perhaps at the point .  We say that f has the limit as z approaches provided that the value f(z) can be made as close as we please to the value by taking z to be sufficiently close to .  When this happens we write

            .  

    The distance between the points z and can be expressed by , so we can give a precise definition similar to the one for a function of two variables.

Definition 2.4 ( limit of f(z) ).  The expression means that for each number , there exists a real number such that  

             whenever .  

Using Equations (1-49) and (1-51), we can also express the last relationship as

             whenever .  

    The formulation of limits in terms of open disks provides a good context for looking at this definition.  It says that for each disk of radius about the point (represented by ) there is a punctured disk of radius about the point (represented by ) such that the image of each point in the punctured -disk lies in the -disk.  The image of the -disk does not have to fill up the entire -disk;  but if z approaches along a curve that ends at , then w=f(z) approaches .  The situation is illustrated in Figure 2.17.

            Figure 2.17   The limit    as  .

Example 2.16.  Show that if  ,  then  ,  where    is any complex number.

Solution.  As f merely reflects points about the y axis, we suspect that any -disk about the point would contain the image of the punctured -disk about    if  .  To confirm this conjecture, we let be any positive number and set  .  Then we suppose that  ,  which means that  .  The modulus of a conjugate is the same as the modulus of the number itself, so the last inequality implies that  .  This is the same as  .  Since   and  ,  this is the same as   ,  which in turn is the same as  , which is what we needed to show.

Explore Solution 2.16.

    If we consider w=f(z) as a mapping from the z plane into the w plane and think about the previous geometric interpretation of a limit, then we are led to conclude that the limit of a function f should be determined by the limits of its real and imaginary parts, u and v.  This conclusion also gives us a tool for computing limits.

Theorem 2.1.  Let    be a complex function that is defined in some neighborhood of  ,  except perhaps at  .  Then  

              

Proof.

Example 2.17.  Show that  .   

Solution.  We have  

              

Computing the limits for u and v, we obtain  

            ,  and
            
            ,  

so our previous theorem implies that .  

Explore Solution 2.17.

    Limits of complex functions are formally the same as those of real functions, and the sum, difference, product, and quotient of functions have limits given by the sum, difference, product, and quotient of the respective limits.  We state this result as a theorem and leave the proof as an exercise.

Theorem 2.2.  Suppose that    and  .  Then  

  

Proof.

Definition 2.5 ( continuity of u(x,y) ).  Let u(x,y) be a real-valued function of the two real variables x and y. We say that u is continuous at the point if the three conditions are satisfied:

  

    Condition (2-23) actually implies Conditions (2-21) and (2-22) because the existence of the quantity on each side of Equation (2-23) is implicitly understood to exist.  For example, if    when    and if  , then    so that Conditions (2-21) , (2-22) , and (2-23) are satisfied.  Hence is continuous at .

    There is a similar definition for complex valued functions.

Definition 2.6 ( continuity of f(z) ).  Let f(z) be a complex function of the complex variable z that is defined for all values of z in some neighborhood of .  We say that f is continuous at if three conditions are satisfied:  

exists,  

Remark 2.3.  Example 2.16 shows that the function ,  is continuous.

    A complex function f is continuous iff its real and imaginary parts, u and v, are continuous.  The proof of this fact is an immediate consequence of Theorem 2.1.  Continuity of complex functions is formally the same as that of real functions, and sums, differences, and products of continuous functions are continuous; their quotient is continuous at points where the denominator is not zero.  These results are summarized by the following theorems.  We leave the proofs as exercises.

Theorem 2.3.  Let    be a defined in some neighborhood of  .  Then    is continuous at    iff    and    are continuous at  .

Proof.

Theorem 2.4.  Suppose that    and    are continuous at the point  .  Then the following functions are continuous at .  

            The sum   ,  

            The difference   ,  

            The product   ,  

            The quotient   ,  provided that  .

            The composition  ,  provided that    is continuous in a neighborhood of the point  .

Proof.

Example 2.18.  Show that the polynomial function given by  

              

is continuous at each point   in the complex plane.

Solution.  If    is the constant function, then  ;  and if  ,  then we can use Definition 2.3 with    and the choice    to prove that    .  Using Property (2-19) and mathematical induction, we obtain  

(2-27)            ,   for   .  

We can extend Property (2-18) to a finite sum of terms and use the result of Equation (2-27) to get   

            .  

Conditions (2-24),  (2-25), and (2-26) are satisfied, so we conclude that P is continuous at .

Explore Solution 2.18.

Extra Example 1.  Show that the polynomial    is continuous at the point    in the complex plane.

Explore Extra Example 1.

        Figure 2.A   The mapping   where  .  

Extra Example 2.  Show that the polynomial    is continuous at the point    in the complex plane.

Explore Extra Example 2.

Extra Example 3.  Show that the polynomial    is continuous at the point    in the complex plane.

Explore Extra Example 3.

    One technique for computing limits is to apply Theorem 2.4 to quotients.  If we let P and Q be polynomials and if  ,  then  

            .  

    Another technique, involves factoring polynomials.  If both and , then P and Q can be factored as
and .   If  , then the limit is  

            .  

Example 2.19.  Show that .  

Solution.  Here P and Q can be factored in the form  

            

so that the limit is obtained by the calculation  

               

Explore Solution 2.19.

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