Math 212 Homework #3 due January 25 Spring 2008
§2.2 Limits and Continuity
Caution: It is NOT enough to prove that a limit is a specific number just by showing that
the limit is that number if you approach it from different directions. Many people made this
mistake on #8(a) and #8(c). Here are the correct solutions.
8a.
lim
(x,y)→(0,0)
(x+y)2−(x−y)2
xy
Multiplying out the top of this fraction, we see that this is the same as
lim
(x,y)→(0,0)
4xy
xy = lim
(x,y)→(0,0) 4=4.
Some people made the mistake of setting one of xor yequal to 0, and then saying the
limit D.N.E. This does not work because x= 0 and/or y= 0 are not in the domain
of the function. (x, y)→(0,0) means that xand yapproach (0,0) along paths in the
domain of the function! It is only the limit point (in our case (0,0)) which is allowed to
be in the boundary of the domain.
8c.
lim
(x,y)→(0,0)
x3−y3
x2+y2
Note that x3−y3= (x−y)(x2+xy +y2). Furthermore, notice that |xy |≤ x2+y2, or
in other words, 0 ≤x2+xy +y2≤2(x2+y2). This leads to the inequality
0≤x2+xy +y2
x2+y2≤2(x2+y2)
x2+y2= 2
We now have the bound
0≤lim
(x,y)→(0,0)
|x3−y3|
x2+y2≤lim
(x,y)→(0,0) |x−y|2 = 0
Therefore, our limit is zero!
Be careful! On the homework, many people saw that this limit was zero in a couple of
experiments (for example set x= 0 or y= 0 or y=x), and decided that the answer
was zero based on these experiments. Even though the answer was correct, this does
not prove it! Also, some people said x2+y2divides x3−y3, which is not true. Some
other people changed the problem to polar coordinates which seemed to work well too.
