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chapter 2 35

# chapter 2 35 - SECTION 2.5 C(2> If f is continuous at a...

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Unformatted text preview: SECTION 2.5 CONTINUITY 99 55. (2>) If f is continuous at a. then by Theorem 8 with g(h) 2 a + h. we have ’13ng +h) 2 f<llliir5(a+ h)) 2 f(a). (<2) Let a > 0. Since lim f(a + h) 2 f(a). there exists 6 > 0 such that 0 < Ihl < 6 2 h—>0 |f(a+h) — f(a)| < e. SoifO < |a: —al < 6. then |f(:I:) — f(a)| 2 |f(a+ (x ~ (1)) — f(a)| < 5. Thus. lirn f(a:) 2 f(a) and so f is continuous at a. I—va 56. lim sin(a + h) 2 lim (sin (1 cos h + cos a sin h) 2 lim (sin (1 cos h) + lim (cos asin h) 11—.0 h_.o h—>0 h—>0 : (’13:?) sina) (A1313 COS h) + ([1152) cos a) (lain) sinh) : (Sin (0(1) + (cos a)(0) 2 sina 57. As in the previous exercise, we must show that lim c0s(a + h) 2 cos a to prove that the cosine function is hﬁrO continuous. . . . lim cos(a + h) 2 11111 (cos (1 cos it — Sln asm h) h—»0 i120 2 lim (cos a cos h) — lim (sin a sin h) h—>O 1220 2 (lim cos (1) (lim cos h) 2 (lim sin a) (lim sin h) h—vO h—bO hAO h—)O 2 (cos a)(1) — (sin a)(0) 2 cosa 58. (a) Since f is continuous at a, lim f(ac) 2 f(a). Thus. using the Constant Multiple Law of Limits. we have E—ba lim (cf )(m) 2 lim cf(ac) 2 c lirn f(m) 2 cf(a) 2 (cf )(a). Therefore cf is continuous at a. (tall xaa (b) Since f and g are continuous at a. lim f(w) 2 ﬂu) and lim g(m) 2 9(a). Since 9(a) 74 0. we can use the IE—>a z—>a lim f(:c) Quotient Law of Limits: lim (i) (x) 2 lim f(;r) 2 17a — f(a) — (i)( ). Thus. i is continuous men 9 05—... 9(rc) 311331 9(1?) 9(a) g 9 at a. 0 if :c is rational ‘ 59. f (ac) 2 is continuous nowhere. For, given any number a and any 6 > 0. the interval 1 if at is irrational (a — 6. a + 6) contains both inﬁnitely many rational and inﬁnitely many irrational numbers. Since f(a) 2 0 or 1. there are inﬁnitely many numbers a: with 0 < (a: — a| < 6 and |f(:c) — f(a)| 2 1. Thus. lim f(:c) 75 f(a). [In fact, lim f (or) does not even exist] 2‘60. 0 if :1: is rational . ‘ 60. g(:c) 2 _ . _ . is continuous at 0. To see why. note that — Jml g g(\$) S |:c]. so by the Squeeze a: ifzc is irrational Theorem linbg(\$) 2 0 2 g(0). But 9 is continuous nowhere else. For ifa 3A 0 and 6 > 0, the interval 1—. (a 2 6. a + 6) contains both inﬁnitely many rational and inﬁnitely many irrational numbers. Since 9(a) 2 0 or a, there are inﬁnitely many numbers m with 0 < la: ~ a] < 5 and Ig(:c) — g(a)| > la! /2. Thus. lim g(;c) # g(a). 61. If there is such a number. it satisﬁes the equation 903 + 1 2 m (2} 3:3 — x + 1 2 0. Let the left—hand side of this equation be called f(a:). Now f(—2) 2 25 < 0. and f(~1) 2 1 > 0. Note also that f(m) is apolynomjal. and thus continuous. So by the Intermediate Value Theorem, there is a number c between —2 and —1 such that f(c) 2 0. so thatc 2 c3 + 1. ...
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