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The Real And Complex Number Systems

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(d) fx 1/1 e 1/x if x 0, f0 0.<br />

Solution: f is continuous on R 0, and since lim x0<br />

e 1/x and lim x0<br />

e 1/x 0,<br />

we know that f has an irremovabel discontinuity at 0. In addition, f0 0and<br />

f0 1, we know that f has the lefthand jump at 0, f0 f0 1, and f is<br />

continuous from the right at 0.<br />

4.59 Locate the points in R 2 at which each of the functions in Exercise 4.11 is not<br />

continuous.<br />

(a) By Exercise 4.11, we know that fx, y is discontinuous at 0, 0, where<br />

fx, y x2 y 2<br />

if x, y 0, 0, andf0, 0 0.<br />

x 2 y2 Let gx, y x 2 y 2 ,andhx, y x 2 y 2 both defined on R 2 0, 0, we know that g<br />

and h are continuous on R 2 0, 0. Note that h 0onR 2 0, 0. Hence, f g/h is<br />

continuous on R 2 0, 0.<br />

(b) By Exercise 4.11, we know that fx, y is discontinuous at 0, 0, where<br />

xy 2<br />

fx, y <br />

if x, y 0, 0, andf0, 0 0.<br />

xy 2 2<br />

x y<br />

Let gx, y xy 2 ,andhx, y xy 2 x y 2 both defined on R 2 0, 0, we know<br />

that g and h are continuous on R 2 0, 0. Note that h 0onR 2 0, 0. Hence,<br />

f g/h is continuous on R 2 0, 0.<br />

(c) By Exercise 4.11, we know that fx, y is continuous at 0, 0, where<br />

fx, y 1 x sinxy if x 0, and f0, y y,<br />

since lim x,y0,0 fx, y 0 f0, 0. Letgx, y 1/x and hx, y sinxy both defined<br />

on R 2 0, 0, we know that g and h are continuous on R 2 0, 0. Note that h 0<br />

on R 2 0, 0. Hence, f g/h is continuous on R 2 0, 0. Hence, f is continuous on<br />

R 2 .<br />

(d) By Exercise 4.11, we know that fx, y is continuous at 0, 0, where<br />

x y sin1/x sin1/y if x 0andy 0,<br />

fx, y <br />

0 if x 0ory 0.<br />

since lim x,y0,0 fx, y 0 f0, 0. It is the same method as in Exercise 4.11, we know<br />

that f is discontinuous at x,0 for x 0andf is discontinuous at 0, y for y 0. <strong>And</strong> it is<br />

clearly that f is continuous at x, y, wherex 0andy 0.<br />

(e) By Exercise 4.11, Since<br />

we rewrite<br />

fx, y <br />

fx, y <br />

cos<br />

sinxsiny<br />

tan xtan y<br />

,iftanx tan y,<br />

cos 3 x if tan x tan y.<br />

xy<br />

cos xcos y<br />

2<br />

cos<br />

xy<br />

2<br />

if tan x tan y<br />

cos 3 x if tan x tan y.<br />

We consider x, y /2, /2 /2, /2, others are similar. Consider two cases (1)<br />

x y, and (2) x y, wehave<br />

(1) (x y) Since lim x,ya,a fx, y cos 3 a fa, a. Hence, we know that f is<br />

,<br />

.

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