Unformatted text preview: 138 10. 11. 12. 13. 14.
15. 16. 17.
18. . False. Consider lim CHAPTER 2 LIMITS AND DERIVATIVES $(z ~ 5) or lim 5111(3: — 5)
z—+5 a: — 5 x—>5 x — 5 . The ﬁrst limit exists and is equal to 5. By Example 3 in Section 2.2. we know that the latter limit exists (and it is equal to 1). 1 II)— False. Consider 3113i} [f(av)g(:c)] = lim [(m — 6) 2—6
. It exists (its value is 1) but f(6) : 0 and 9(6) does not exist. so f(6)g(6) 75 1. . True. A polynomial is continuous everywhere. so lin‘iptr) exists and is equal to p(b).
. . . 1 1 . . . . . . . .
. False. Constder 1111i) [f(m) i g(a:)] : 11116 52— — E31 . This llmrt IS —00 (not 0), but each of the 1nd1v1dual functions approaches 00. . True. See Figure 4 in Section 2.6. False. Consider f (as) = sina: for :e Z 0. lim f (m) ¢ :00 and f has no horizontal asymptote. z—>oo 1 ac — 1 if m 1
False. Consider f(:1:) : /( ) 76
2 if a: : 1 False. The function f must be continuous in order to use the Intermediate Value Theorem. For example, let 1 if 0 g at < 3
f(g;) : 1 'f 3 There is no number 0 E [07 3] with f(c) = 0.
— 1 a: = True. Use Theorem 2.5.8 with a = 2, b : 5. and 9(1):) : 4:02 — 11. Note that f(4) = 3 is not needed. True. Use the Intermediate Value Theorem with a : —1. b = 1, and N : 7r, since 3 < 7r < 4. True, by the deﬁnition of a limit with a = 1. 3:2 + 1 if a: 55 0 False. For example, let f (at) = .
2 if m : 0 Then f(m) > 1 for all :17. but lirq) f(m) : lim (m2 + 1) 2 1. 23—0 False. See the note after Theorem 4 in Section 2.9. True. f’(r) exists :> f is differentiable atr :> f is continuous at 7’ => lim f(;v) : f0"). Ct—VI‘ /—’—— EXERCISES ———————’ 1. ' ' : " l' : 0
(a) (1) $113+ f(:r:) 3 (n) £3113 + f (:6)
(iii) lirn3 f (1;) does not exist since the left and right (iv) 33: ﬂan) : 2 limits are not equal. (The left limit is —2.) (v) lim ﬂat) 2 00 (vi) lim f(a7) : 700
$—’0 maZ’
(vii) lim f(a:) : 4 (viii) lim ﬂan) : —1
(II—>00 {84*00
(b) The equations of the horizontal asymptotes are y = #1 and y : 4. (c) The equations of the vertical asymptotes are a: : 0 and w = 2.
(d) f is discontinuous at m z —3, 0, 2, and 4. The discontinuities are jump, infinite, inﬁnite, and removable, respectively. ...
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 Spring '10
 Ban
 Calculus

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