四 4. 泰勒定理的應用

四 _4. 泰勒定理的應用

4.2 (1)

f⁰(x) = −sinx + x, f⁰(0) = 0 f⁰⁰(x) = −cosx + 1, f⁰⁰(0) = 0 f⁰⁰⁰(x) = sinx, f⁰⁰⁰(0) = 0 f⁽⁴⁾ = cosx, f⁽⁴⁾(0) = 1 > 0 由定理_4.1知_{,f (0) = 0}是極小值 (2)

因為_{x = 0}是_f⁰_{(x) = 0}唯一的根_, 所以_{f (0) = 0}也是最小值_, 因此_{f (x) ≥ 0}

4.4 (2)lim

x→0

x−tan⁻¹x

x³ = lim

x→0 1− ¹

1+x2

3x² = lim

x→0 x²

3x²(1+x²) = lim

x→0 1

3(1+x²) = ¹₃

(5)lim

x→0 7^x−1

2^x−1 = lim

x→0 7^xln 7 2^xln 2 = ^{ln 7}_{ln 2}

(7)lim

x→0 x²

ln sec x = lim

x→0

2x

1

sec xsec x tan x = lim

x→0 2x

tan x = lim

x→0 2 sec²x = 2

(8)lim

x→0 e^{sin x}−1

sin x = lim

x→0

cos xe^{sin x}

cos x = lim

x→0e^{sin x}= 1

4.5 (1) lim

x→∞

log₃x

ln(x²+1) = lim

x→∞

1 ln 3

1 x 1

x2+12x = lim

x→∞

x²+1

2(ln 3)x² = lim

x→∞

2x

4(ln 3)x = lim

x→∞

1

2 ln 3 = _{2 ln 3}¹

(3) lim

x→∞

x²+x+1

e^x = lim

x→∞

2x+1

e^x = lim

x→∞

2 e^x = 0

(4) lim

x→∞

ln³x

x = lim

x→∞

3 ln²x·_x¹

1 = lim

x→∞

3 ln²x

x = lim

x→∞

6 ln x·_x¹

1 = lim

x→∞

6 ln x

x = lim

x→∞

6·¹_x 1

= lim

x→∞

6 x = 0

4.6 (2)

x→0lim(_x¹2 − _{sin x}¹ ) = lim

x→0

sin x−x²

x²sin x = lim

x→0

cos x−2x

2x sin x+x²cos x = ∞

(5)

x→0limx^x = lim

x→0e^{x ln x} = e^{( limx→0x ln x)}

x→0limx ln x = lim

x→0 ln x

1 x

= lim

x→0

1 x

−x⁻² = lim

x→0−x = 0 因此_lim

x→0x^x = e⁰ = 1

1

(6)

x→∞lim(1 + x)^x1 = lim

x→∞e^x1ln(1+x) = e^{( limx→∞}

1

xln(1+x))

x→∞lim

1

xln(1 + x) = lim

x→∞

ln(1+x)

x = lim

x→∞

1 1+x

1 = lim

x→∞

1 1+x = 0 因此_lim

x→∞(1 + x)^x1 = e⁰ = 1

2