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.

Enter the function  .  

Find the iterated limit,  .  

Find the iterated limit,  .  

Find the polar limit  .  

So, along all lines through the origin, the limit is 0.

We see that the limit should be 0.

A more rigorous proof is needed for other approaches to the origin.  

                 

We can use Mathematica to make a ContourPlot of  u(x,u).

(c) 2006 John H. Mathews, Russell W. Howell