If f (x) and g(x) are polynomials and the degree of g is smaller than the degree of f , then in .NET

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9.1 Evaluate the following limits: (a) (c)
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x + x
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If f (x) is a polynomial,
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x
f (x) = an x n + an 1 x n 1 + an 2 x n 2 + + a1 x + a0
2 The
general rule for determining whether + or holds is complex. If an and bk are the leading coef cients of f and g, respectively, then the an n k limit is equal to lim x and the correct sign is the sign of an bk ( 1)n k . x bk
SPECIAL LIMITS
[CHAP. 9
with an = 0, then a an 1 /an a /an a /an f (x) /an + n 2 + + 1 + 0n n =1+ 2 n 1 an x x x x x It follows from (9.1) that as x , f (x)/an x n becomes arbitrarily close to 1. Therefore, f (x) must become unbounded exactly as does an x n ; that is,
x
lim f (x) =
x
lim an x n
Applying this rule to the given polynomials, we nd: (a) (c)
x + x
lim 5x 3 = +
(b) (d)
x +
lim ( 2x 3 ) = lim 7x 3 =
lim 7x 4 = +
x
GENERAL RULE P. If f (x) = an xn + an 1 xn 1 + + a1 x + a0 , then:
(i) (ii)
x + x
lim f (x) = and the correct sign is the sign of an .
lim f (x) = and the correct sign is the sign of an ( 1)n .
9.2 Evaluate:
(a) lim
x 1+
x+3 x+3 ; (b) lim . x 1 x 1 x 1
x+3 increases without x 1
As x approaches 1, the denominator x 1 approaches 0, whereas x + 3 approaches 4. Thus, bound.
(a) As x approaches 1 from the right, x 1 is positive. Since x + 3 is positive when x is close to 1, x+3 >0 x 1 and
x 1+ x 1
= +
(b) As x approaches 1 from the left, x 1 is negative, while x + 3 is positive. Therefore, x+3 <0 x 1 and lim x+3 = x 1
x 1
The line x = 1 is a vertical asymptote of the graph of the rational function (see Fig. 9-7).
9.3 Given f (x) =
1/x 3x + 2
if x > 0 nd: if x < 0
(a) lim f (x); (b) lim f (x); (c) lim f (x); (d) lim f (x).
x 0+ x 0 x + x
x 0+
lim f (x) = lim
x 0+ x
= +
Hence, the line x = 0 (the y-axis) is a vertical asymptote of the graph of f (Fig. 9-8). (b) lim f (x) = lim (3x + 2) = 2
x 0
x 0
x +
lim f (x) =
x + x
Hence, the line y = 0 (the x-axis) is a horizontal asymptote (to the right) of the graph of f . (d) lim f (x) = lim (3x + 2) =
x
x
CHAP. 9]
SPECIAL LIMITS
Fig. 9-7
Fig. 9-8
9.4 Evaluate:
(a) lim
3x 2 5 2x 7 5x + 2 5x + 2 ; (d) lim . ; (b) lim 2 ; (c) lim x 4x 2 + 5 x + x 8 x + x 2 3x + 1 x x 2 3x + 1
(a) Apply Rule B of Section 9.3: 3 3x 2 5 = x 4x 2 + 5 4 lim (b) Apply Rule A of Section 9.3: lim 2x 7 =0 x 2 as x + . But x 2 = x when x > 0. Thus,
x + x 2 8
(c) By Rule P, developed in Problem 9.1, the denominator behaves like by Rule B of Section 9.3, lim 5x + 2 x 2 3x + 1 = lim
x +
5x + 2 5 = =5 x + x 1 x 2 as x . But x 2 = x when x < 0.
(d) By Rule P, developed in Problem 9.1, the denominator behaves like Thus, by Rule B of Section 9.3, 5x + 2 x 2 3x + 1 =
x
x
5 5x + 2 = = 5 x 1
9.5 Find the vertical and horizontal asymptotes of the graph of the rational function f (x) = 3x 2 5x + 2 6x 2 5x + 1
SPECIAL LIMITS
[CHAP. 9
The vertical asymptotes are determined by the roots of the denominator: 6x 2 5x + 1 = 0 (3x 1)(2x 1) = 0 3x 1 = 0 3x = 1 x= 1 3 or or or 2x 1 = 0 2x = 1 x= 1 2
Because the numerator is not 0 at x = 1 or x = 1 , | f (x)| approaches + as x approaches 1 or 1 from one side or the other. 3 2 3 2 Thus, the vertical asymptotes are x = 1 and x = 1 . 3 2 The horizontal asymptotes may be found by Rule B of Section 9.3, 3x 2 5x + 2 3 1 = = 2 5x + 1 x + 6x 6 2 lim Thus, y = 1 is a horizontal asymptote to the right. A similar procedure shows that lim f (x) = 1 , and so y = 1 is also a 2 2 2 x horizontal asymptote to the left.
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