Unformatted text preview: 82 CHAPTER 2 LIMITS AND DERIVATIVES 51. The graph of f($) : [[wll + ﬂeml] is the same as the graph of g(:c) = —1 with holes at each integer. since f(a) = 0 for any integer a. Thus. In? f(a:) = ,1 and lim+ f (cc) 2 —1. so lirr12f(x) = —1. However.
10—» ’ ac—>2 13'” f(2) = [[2]] + I—2I = 2 + (—2) = o. 53W) aé W) 2
. v
52. 11m (Lo 1 —— # > = Lox/1 — : 0. As the velocity approaches the speed of light, the length approaches 0. u—n" C2 A leftehand limit is necessary since L is not deﬁned for v > c.
53. Since p(:n) is a polynomial, p(:c) : a0 + mm + a2m2 +  ‘ . + anx". Thus. by the Limit Laws, lim 12(23): lirn (a0 + (113: + (12.272 + ~  ‘ + aux") I—Nl 519—90. :a0+a1lim$+azlimm2+~~+anlimmn (Ball (II—>a {E—>a zao+a1a+a2a2+~~+anan=p(a) Thus. for any polynomial p. lim p(a:) = p(a). z—m. 54. Let Mat) 2 p_(:c_) where p(:r:) and q(m) are any polynomials. and suppose that q(a) 75 0. Thus, (1(33)
PW) lim p(m) (a)
$13: Mm) : align1 R7?) : W [Limit Law 5] : 2% [Exercise 53] : r(a). :E—>a 55. Observe that O 3 ﬂap) 3 1:2 for all :0. and lim 0 : 0 : lim 272. So, by the Squeeze Theorem, lim0 f($) : 0. 1—»0 m—rO 56. Let ﬂag) 2 [[20]] and g(x) : —[[ac]]. Then lim f($) and lim g(m) do not exist (Example 10) but z—>3 :c——>3 £313 If(93) + 9(mll = $1313 (IIUBII  IlirII) : 11m 0 = 0 14>?) 57. Let f(m) : H(a:) and 9(31) : 1 — H(ac) where H is the Heaviside function deﬁned in Exercise 1.3.59.
Thus. either f or g is 0 for any value of 2:. Then limO f (m) and lira] g(m) do not exist. but lirn [f(a:)g(m)] : lim 0 : 0. z—rO 33—4) «317—2 (WT—7572 \/6_—a:+2 Vii—3+1)
I 58.1' —————:1'  
zlgéx/IS—mel $1312 x/3—cc1 x/6—ac+2 x/3#w+1 59. Since the denominator approaches 0 as :c —> —2. the limit will exist only if the numerator also approaches 0 as
m —+ ,2. In order for this to happen. we need lim2 (3x2 + a1: + a + 3) : 0 <1) :c—>; 3(—2)2+a(!2)+a+320 4:) 12—2a+a+3=0 41> a:15.Witha:15.thelimitbecomes
3m2+15x+18 f 1. 3(m+2)($+3) _ lim 3(x+3) : 3(#2+3) lim 3
_ :—:—1.
z 2 (ﬁlm; 2 351311295 1)(:c+2) 1+2 33—1 —2—1 —3 ...
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 Calculus

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