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3个回答
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(2)代入,为0.
(4)用洛必达法则,为1/2.
(6)原式-->(x-2)/(x-1)-->2/3.
(8)2.
(10)分子有理化:原式-->4/[√(5x-4)+√x]
-->2.
(4)用洛必达法则,为1/2.
(6)原式-->(x-2)/(x-1)-->2/3.
(8)2.
(10)分子有理化:原式-->4/[√(5x-4)+√x]
-->2.
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(2)当x趋于2时,
lim(x²-4)/x+5
=0
(4)当x趋于0时,
lim(4x³-2x²+x)/3x²+2x
=lim(4x²-2x+1)/3x+2
=1/2
(6)当x趋于4时,
lim(x²-6x+8)/x²-5x+4
=lim(x-2)(x-4)/(x-1)(x-4)
=lim(x-2)/(x-1)
=2/3
(8)当x趋于无穷时,
lim(1+1/x)/(2-1/x²)
=1/2
(10)当x趋于1时,
lim(√5x-4-√x)/x-1
=lim(√5x-4-√x)(√5x-4+√x)/(x-1)(√5x-4+√x)
=lim(5x-4-x)/(x-1)(√5x-4+√x)
=lim4(x-1)/(x-1)(√5x-4+√x)
=lim4/(√5x-4+√x)
=2
lim(x²-4)/x+5
=0
(4)当x趋于0时,
lim(4x³-2x²+x)/3x²+2x
=lim(4x²-2x+1)/3x+2
=1/2
(6)当x趋于4时,
lim(x²-6x+8)/x²-5x+4
=lim(x-2)(x-4)/(x-1)(x-4)
=lim(x-2)/(x-1)
=2/3
(8)当x趋于无穷时,
lim(1+1/x)/(2-1/x²)
=1/2
(10)当x趋于1时,
lim(√5x-4-√x)/x-1
=lim(√5x-4-√x)(√5x-4+√x)/(x-1)(√5x-4+√x)
=lim(5x-4-x)/(x-1)(√5x-4+√x)
=lim4(x-1)/(x-1)(√5x-4+√x)
=lim4/(√5x-4+√x)
=2
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