ASSIGNMENT 5 SOLUTION
JAMES MCIVOR
1. Stewart 14.2.16
[5 pts] Find the limit or say why it does not exist:
x2 sin2 y
(x,y)→(0,0) x2 + 2y 2
lim
Solution: The limit is equal to zero. To see this, use the Squeeze Theorem. Since x2 ≤ x2 + 2y 2 ,
x2
we have x2 +2y
2 ≤ 1, therefore
2 2 x sin y 2 ≤ sin y 0 ≤ 2
x + 2y 2 Since sin2 y goes to zero as x, y go to zero, the middle term does also.
2. Stewart 14.2.38
[5 pts] Determine the set of points at which the function
(
xy
if (x, y) 6= (0, 0)
2
2
f (x, y) = x +xy+y
1
if (x, y) = (0, 0)
is discontinuous.
xy
Solution: The function x2 +xy+y
2 is continuous on its domain. The domain consists of the points
where the denominator is nonzero. We can find these points by solving x2 +xy +y 2 = 0. Completing
the square for x gives (x + 12 y)2 + 43 y 2 = 0, and this happens only when (x, y) = (0, 0). So f is
indeed continuous away from the origin, i.e., the only possibile discontinuity is at the origin. [Note:
It was not necessary to explicitly determine the domain to get full points - you could
just take Stewart’s word for it] To see whether f is continuous there, we find
lim
f (x, y)
(x,y)→(0,0)
As we approach the origin along the line x = 0, the limit is zero, but as we approach the origin
along the line y = x, the limit is 1/3. Thus the above limit does not exist, so f is not continuous at
the origin.
3. Stewart 14.2.44
[5 pts - NOTE - You can assume that a is a rational number.] Let
(
0 if y ≤ 0 or y ≥ x4
f (x, y) =
1 if 0 < y < x4
(a) Show that f (x, y) → 0 as (x, y) → (0, 0) along any path of the form y = mxa , where a < 4.
(b) Despite part (a), show that f is discontinuous at the origin.
(c) Show that f is discontinuous on two entire curves.
Solution:
(a) First consider the limit along the curve y = mxa as x goes to zero from above (i.e., x > 0). If
m is negative, then y < 0 along this curve, so f (x, y) is always 0 along the curve, so the limit is
0. If m is positive, then as soon as x < m1/(4−a) , we have mxa > x4 , so f (x, y) = 0 along this
curve, once we get sufficiently close to the origin (i.e., once x < m1/(4−a) ). Thus the limit along
this curve is 0.
Now let’s consider approaching the origin along these same curves, but from the left, when
x < 0. First assume m ≥ 0. Under the assumption that a is rational, write a = p/q, where p
1
2
JAMES MCIVOR
and q are not both even (if they are, we can just cancel some twos). Then mxa = m(x1/q )p . We
consider various cases:
(1) q is even. Then the expression mxa is undefined, so there is no curve for x < 0.
(2) q and p are both odd. Then x1/q < 0, so mxa < 0, and f (x, y) is always zero along this
curve, so the limit along the curve is zero.
(3) q is odd but p is even. Here x1/q < 0 but mxa is positive. The same argument as in the
previous paragraph shows that once x is close to zero, f (x, y) = 0 along the curve, so this
limit is zero.
Finally, suppose instead that m were negative. Then the two cases (2) and (3) above are
reversed, but the same arguments go through.
(b) f is discontinuous at (0, 0) since along the curve y = x5 , for instance, f (x, y) = 1, so as we
approach the origin along this curve, the limit is one. Hence
lim
f (x, y)
(x,y)→(0,0)
does not exist, since it depends on the curve of approach.
(c) f is discontinuous at every point along each of the curves y = 0 and y = x4 . This is because
the values of f are zero on one side of the curve, and 1 on the other. To be more precise, let
(a, a4 ) be a point along the curve y = x4 , with a > 0. Then as x → a from the right along the
horizontal line y = a4 , f (a, a4 ) = 1, but as we approach the point (a, a4 ) along that sam line as
x → a from the left, we have f (x, y) = 0. Thus this limit does not exist. Similarly when a < 0.
A similar argument can be made for points along the x-axis.