题目
高数【函数极限】
lim (1+1/x)=?
x→∞
lim(1-3x)∧1/(2x)=?
x→0
lim[(2x+3)/(2x+1)]∧(x+1)
x→∞
lim (1+1/x)=?
x→∞
lim(1-3x)∧1/(2x)=?
x→0
lim[(2x+3)/(2x+1)]∧(x+1)
x→∞
提问时间:2021-03-18
答案
1.lim (1+1/x)=1+0=1
x→∞
2.lim(1-3x)∧1/(2x)
x→0
=lim(1-3x)^[(-1/3x)*(-3/2)]
x→0
=lim[(1-3x)^(-1/3x)]^(-3/2)
x→0
=e^(-3/2)
3.lim[(2x+3)/(2x+1)]∧(x+1)
x→∞
=lim[1+1/(x+1/2)]^[(x+1/2)+1/2]
x→∞
=lim[1+1/(x+1/2)]^(x+1/2)*[1+1/(x+1/2)]^(1/2)
x→∞
=e*(1+0)^(1/2)
=e
x→∞
2.lim(1-3x)∧1/(2x)
x→0
=lim(1-3x)^[(-1/3x)*(-3/2)]
x→0
=lim[(1-3x)^(-1/3x)]^(-3/2)
x→0
=e^(-3/2)
3.lim[(2x+3)/(2x+1)]∧(x+1)
x→∞
=lim[1+1/(x+1/2)]^[(x+1/2)+1/2]
x→∞
=lim[1+1/(x+1/2)]^(x+1/2)*[1+1/(x+1/2)]^(1/2)
x→∞
=e*(1+0)^(1/2)
=e
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