高数怎么做？
答案:2 悬赏:20 手机版
解决时间 2021-01-23 21:41
- 提问者网友：精神病院里
- 2021-01-23 15:25
高数怎么做？
最佳答案
- 五星知识达人网友：不想翻身的咸鱼
- 2021-01-23 15:38
(1)
x->0
tanx ~ x+(1/3)x^3
sinx ~ x-(1/6)x^3
tanx -sinx ~(1/2)x^3
lim(x->0) ( tanx- sinx)/(sinx)^3
=lim(x->0) (1/2)x^3/x^3
=1/2
(2)
分子
sinx - tanx ~ -(1/2)x^3
分母
(1+x^2)^(1/3) ~ 1+(1/3)x^2
(1+x^2)^(1/3) -1 ~ (1/3)x^2
√(1+sinx) ~ √(1+x) ~ 1+(1/2)x
√(1+sinx) -1 ~ (1/2)x
lim(x->0) (sinx- tanx) /{ (1+x^2)^(1/3) . [√(1+sinx) -1] }
= lim(x->0) -(1/2)x^3 /{ (1/3)x^2 . (1/2)x }
=-3
(3)
lim(x->1) [ 1/(x-1) - (x+2)/(x^3 -1) ]
=lim(x->1) [(x^2+x+1) -(x+2) ] /[(x-1)(x^2+x+1)]
=lim(x->1) (x^2-1) /[(x-1)(x^2+x+1)]
=lim(x->1) (x+1) /(x^2+x+1)
=(1+1)/(1+1+1)
=2/3
(4)
lim(x->+∞) ) (3x^2 + x +2 )/( x^3+2x-1)
=lim(x->+∞) ) 3x^2/ x^3
=lim(x->+∞) ) 3/x
=0
x->0
tanx ~ x+(1/3)x^3
sinx ~ x-(1/6)x^3
tanx -sinx ~(1/2)x^3
lim(x->0) ( tanx- sinx)/(sinx)^3
=lim(x->0) (1/2)x^3/x^3
=1/2
(2)
分子
sinx - tanx ~ -(1/2)x^3
分母
(1+x^2)^(1/3) ~ 1+(1/3)x^2
(1+x^2)^(1/3) -1 ~ (1/3)x^2
√(1+sinx) ~ √(1+x) ~ 1+(1/2)x
√(1+sinx) -1 ~ (1/2)x
lim(x->0) (sinx- tanx) /{ (1+x^2)^(1/3) . [√(1+sinx) -1] }
= lim(x->0) -(1/2)x^3 /{ (1/3)x^2 . (1/2)x }
=-3
(3)
lim(x->1) [ 1/(x-1) - (x+2)/(x^3 -1) ]
=lim(x->1) [(x^2+x+1) -(x+2) ] /[(x-1)(x^2+x+1)]
=lim(x->1) (x^2-1) /[(x-1)(x^2+x+1)]
=lim(x->1) (x+1) /(x^2+x+1)
=(1+1)/(1+1+1)
=2/3
(4)
lim(x->+∞) ) (3x^2 + x +2 )/( x^3+2x-1)
=lim(x->+∞) ) 3x^2/ x^3
=lim(x->+∞) ) 3/x
=0
全部回答
- 1楼网友：青灯有味
- 2021-01-23 16:33
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