ASSIGNMENT 5 SOLUTION
JAMES MCIVOR
1. Stewart 14.2.16 [5 pts]Find the limit or say why it does not exist:
lim
(x,y)→(0,0)
x2sin2y x2+ 2y2
Solution: The limit is equal to zero. To see this, use the Squeeze Theorem. Sincex2≤x2+ 2y2, we have x2+2yx2 2 ≤1, therefore
0≤
x2sin2y x2+ 2y2
≤ sin2y
Since sin2y goes to zero asx, ygo to zero, the middle term does also.
2. Stewart 14.2.38 [5 pts]Determine the set of points at which the function
f(x, y) =
( xy
x2+xy+y2 if (x, y)6= (0,0) 1 if (x, y) = (0,0) is discontinuous.
Solution: The function x2+xy+yxy 2 is continuous on its domain. The domain consists of the points where the denominator is nonzero. We can find these points by solvingx2+xy+y2= 0. Completing the square for x gives (x+12y)2+34y2 = 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
(x,y)→(0,0)f(x, y)
As we approach the origin along the linex= 0, the limit is zero, but as we approach the origin along the liney=x, the limit is 1/3. Thus the above limit does not exist, sof is not continuous at the origin.
3. Stewart 14.2.44
[5 pts - NOTE - You can assume that a is a rational number.] Let f(x, y) =
(0 if y≤0or y≥x4 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, wherea <4.
(b) Despite part (a), show thatf is discontinuous at the origin.
(c) Show thatf is discontinuous on two entire curves.
Solution:
(a) First consider the limit along the curve y=mxa as xgoes to zero from above (i.e.,x >0). If mis negative, theny <0 along this curve, sof(x, y) is always 0 along the curve, so the limit is 0. Ifmis 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., oncex < 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 assumem ≥0. Under the assumption that a is rational, write a=p/q, wherep
1
2 JAMES MCIVOR
andqare not both even (if they are, we can just cancel some twos). Thenmxa=m(x1/q)p. We consider various cases:
(1) qis even. Then the expressionmxa is undefined, so there is no curve forx <0.
(2) q and p are both odd. Then x1/q <0, so mxa < 0, andf(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 xis 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
(x,y)→(0,0)lim f(x, y) does not exist, since it depends on the curve of approach.
(c) f is discontinuous at every point along each of the curvesy = 0 and y =x4. This is because the values off 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 curvey=x4, witha >0. Then asx→afrom the right along the horizontal liney=a4,f(a, a4) = 1, but as we approach the point (a, a4) along that sam line as x→afrom the left, we havef(x, y) = 0. Thus this limit does not exist. Similarly whena <0.
A similar argument can be made for points along thex-axis.
