麻烦帮忙解一下微积分题,要具体解题过程和原理 第一道:lim(((1-x)^1/2)-3)/(2+x^1/3) x趋于-8
2个回答
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第一题:
lim(x→-8) [√(1 - x) - 3]/[2 + x^(1/3)]
= lim(x→-8) [√(1 - x) - 3]/[2 + x^(1/3)] · [√(1 - x) + 3]/[√(1 - x) + 3] <==分子有理化
= lim(x→-8) [(1 - x) - 9]/[(2 + x^(1/3))(√(1 - x) + 3)]
= lim(x→-8) - (8 + x)/[2 + x^(1/3)]
= lim(x→-8) - [(2 + x^(1/3))(4 - 2x^(1/3) + x^(2/3))]/[(2 + x^(1/3))(√(1 - x) + 3)] <==分子因式分解
= lim(x→-8) - [4 - 2x^(1/3) + x^(2/3)]/(√(1 - x) + 3) <==约掉2 + x^(1/3)
= - [4 - 2(- 8)^(1/3) + (- 8)^(2/3)]/[√(1 + 8) + 3]
= - [4 - 2(- 2) + 4]/6
= - 2
第二题:
lim(x→-1) (4 + x + x³ + x⁵ +x⁷)/(x + 1)
= lim(x→-1) [(x + 1)(x⁶ - x⁵ + 2x⁴ - 2x³ + 3x⁵ - 3x + 4)]/(x + 1) <==多项式除法,约掉x + 1
= lim(x→-1) (x⁶ - x⁵ + 2x⁴ - 2x³ + 3x² - 3x + 4)
= (- 1)⁶ - (- 1)⁵ + 2(- 1)⁴ - 2(- 1)³ + 3(- 1)² - 3(- 1) + 4
= 1 + 1 + 2 + 2 + 3 + 3 + 4
= 16
lim(x→-8) [√(1 - x) - 3]/[2 + x^(1/3)]
= lim(x→-8) [√(1 - x) - 3]/[2 + x^(1/3)] · [√(1 - x) + 3]/[√(1 - x) + 3] <==分子有理化
= lim(x→-8) [(1 - x) - 9]/[(2 + x^(1/3))(√(1 - x) + 3)]
= lim(x→-8) - (8 + x)/[2 + x^(1/3)]
= lim(x→-8) - [(2 + x^(1/3))(4 - 2x^(1/3) + x^(2/3))]/[(2 + x^(1/3))(√(1 - x) + 3)] <==分子因式分解
= lim(x→-8) - [4 - 2x^(1/3) + x^(2/3)]/(√(1 - x) + 3) <==约掉2 + x^(1/3)
= - [4 - 2(- 8)^(1/3) + (- 8)^(2/3)]/[√(1 + 8) + 3]
= - [4 - 2(- 2) + 4]/6
= - 2
第二题:
lim(x→-1) (4 + x + x³ + x⁵ +x⁷)/(x + 1)
= lim(x→-1) [(x + 1)(x⁶ - x⁵ + 2x⁴ - 2x³ + 3x⁵ - 3x + 4)]/(x + 1) <==多项式除法,约掉x + 1
= lim(x→-1) (x⁶ - x⁵ + 2x⁴ - 2x³ + 3x² - 3x + 4)
= (- 1)⁶ - (- 1)⁵ + 2(- 1)⁴ - 2(- 1)³ + 3(- 1)² - 3(- 1) + 4
= 1 + 1 + 2 + 2 + 3 + 3 + 4
= 16
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