求极限，第17和第18题
答案:1 悬赏:60 手机版
解决时间 2021-04-01 18:43
- 提问者网友：呐年旧曙光
- 2021-04-01 04:49
求极限，第17和第18题
最佳答案
- 五星知识达人网友：逃夭
- 2021-04-01 04:58
17.
lim [1/(x+1) -3/(x³+1)]
x→-1
=lim [(x²-x+1)/(x+1)(x²-x+1) -3/(x+1)(x²-x+1)]
x→-1
=lim [(x²-x+1-3)/(x+1)(x²-x+1)]
x→-1
=lim [(x²-x-2)/(x+1)(x²-x+1)]
x→-1
=lim [(x+1)(x-2)/(x+1)(x²-x+1)]
x→-1
=lim [(x-2)/(x²-x+1)]
x→-1
=(-1-2)/[(-1)²-(-1)+1]
=-1
18.
lim x[√(x²+1)-x]
x→+∞
=lim x[√(x²+1)-x][√(x²+1)+x]/[√(x²+1)+x]
x→+∞
=lim x[(x²+1)-x²]/[√(x²+1)+x]
x→+∞
=lim x/[√(x²+1)+x]
x→+∞
=lim 1/[√(1+ 1/x²)+1]
x→+∞
=1/(√1+1)
=1/(1+1)
=½追问
第9题可以教一下我吗？
lim [1/(x+1) -3/(x³+1)]
x→-1
=lim [(x²-x+1)/(x+1)(x²-x+1) -3/(x+1)(x²-x+1)]
x→-1
=lim [(x²-x+1-3)/(x+1)(x²-x+1)]
x→-1
=lim [(x²-x-2)/(x+1)(x²-x+1)]
x→-1
=lim [(x+1)(x-2)/(x+1)(x²-x+1)]
x→-1
=lim [(x-2)/(x²-x+1)]
x→-1
=(-1-2)/[(-1)²-(-1)+1]
=-1
18.
lim x[√(x²+1)-x]
x→+∞
=lim x[√(x²+1)-x][√(x²+1)+x]/[√(x²+1)+x]
x→+∞
=lim x[(x²+1)-x²]/[√(x²+1)+x]
x→+∞
=lim x/[√(x²+1)+x]
x→+∞
=lim 1/[√(1+ 1/x²)+1]
x→+∞
=1/(√1+1)
=1/(1+1)
=½追问
第9题可以教一下我吗？
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