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chapter 2 29

# chapter 2 29 - SECTION 2.5 CONTINUITY 93 Since it is given...

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Unformatted text preview: SECTION 2.5 CONTINUITY 93 Since it is given that lim [2f(av) — g(:c)] 2 4. we have 10 — 9(3) 2 4. so 9(3) 2 6. z—v3 10.1im f(m):1in}1(m2 + x/7 — m) = liaise? + /lin}17— lirrila: : 42 + v/7— 4 =16 + ¢§ = M)- z—>4 m—> 39—» x—v (2., By the deﬁnition of continuity, f is continuous at a = 4. 4 11. lim m) = lim (m+2:c3)4 = ( lim w+2 lim \$3) —[ 1+2( 1)3]4 —(—3)4_81_f(—1). x—i—l z—i—l m—r—l m—v—l By the deﬁnition of continuity. f is continuous at a = —1. lim LE + lim 1 : :ﬁ—=———=—: 4.5 ‘t 514. 12' 11319”) 11331239 *1 2 lirrilm2 — 1111311 2(4)2 —1 31 g( ) 0915“)“ "mo“ a 217+?) lim(2a:+3) 2limx+ lim3 . ~ . : - : i—‘Fa L. .t 5 : (.BQCL fan 13 Fora > 2 we have £1111 f(a:) 1113i; 95 _ 2 ——lim(w ¥ 2) [ 1111] Law ] 511m \$ _ 11m 2 2a + 3 . . . . , [1. 2. and 3] : 2 [7 and 8] : f(a). Thus, f 18 continuous atcc : a for every am (2, 00): that is. f IS a 2 continuous on (2. oo). 14. For a < 3. we have lim g(x) : lim 2 V3 — a: = 211m V3 — 3: [Limit Law 3] : 2 lim (3 — \$) [1 1] I‘M]. m—va 13.40. Iﬂa : 2 lim 3 — lim 2: [2] : 2 V3 — a [7 and 8] = g(a). so 9 is continuous at a: = a for every a in z—ta ilk—ta. (~00, 3). Also. lim 9(95) : 0 : 9(3). so g is continuous from the left at 3. Thus. 9 is continuous on (—00. 3]. z—>3’ 15. f(ac) : 111 [91: ~ 2| is discontinuous at 2 since f(2) : an is not deﬁned. 1/(3: — 1) if x 7E 1 16. ﬂat) : 2 'f 1 is discontinuous at 1 because I a: = lim1 f (x) does not exist. z—> SE 6 ifac<0 17. f(a:) :{ 3:2 ifoO The left—hand limit off at a : 0 is lim f(:r) 2 lim em : 1. The m—)O* zaO‘ right—hand limit off at a : 0 is lim f(m) = lim \$2 : 0. Since m—>O+ w—>O+ these limits are not equal. lirn) f(ac) does not exist and f is z—> discontinuous at 0. ...
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