ASSIGNMENT 5 SOLUTION

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

lim

(x,y)(0,0)

x2

+

2y2

Solution: The limit is equal to zero. To see this, use the Squeeze Theorem. Since x2 x2 + 2y2,

we

have

x2 x2 +2y 2

1,

therefore

0

x2 sin2 y x2 + 2y2

sin2 y

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

is discontinuous.

xy

f (x, y) = x2+xy+y2 1

if (x, y) = (0, 0) if (x, y) = (0, 0)

Solution:

The

function

xy 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 +y2 = 0. Completing

the

square

for

x

gives

(x +

1 2

y)2

+

3 4

y2

=

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.

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