求解极限!!
2个回答
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(1)
lim<n→∞>(n/2)R²sin(2π/n)极限不存在
(3)
lim<x→0>arctan3x/sin2x
=lim<x→0>(arctan3x)'/(sin2x)'
=lim<x→0>[3/(1+9x²)]/(2cos2x)
=[3/(1+0)]/(2×1)
=3/2
(5)
lim<x→0>(1-cos4x)/(xsinx)
=lim<x→0>[1-(1-2sin²2x)]/(xsinx)
=lim<x→0>2sin²2x/(xsinx)
=lim<x→0>2(2sinxcosx)²/(xsinx)
=lim<x→0>8sin²xcos²x/(xsinx)
=lim<x→0>8sinx/x
=8
(1)
lim<x→∞>[x/(1+x)]^(x-3)
=lim<x→∞>[1-(1/1+x)]^(x-3)
=lim<x→∞>{[1+(-1/1+x)]^-(1+x)}^[(x-3)/-(1+x)]
=e^lim<x→∞>[(-x+3)/(x+1)]
=e^(-1)
=1/e
(2)
lim<x→∞>[(2x+1)/(2x-1)]^x
=lim<x→∞>[(2x-1+2)/(2x-1)]^x
=lim<x→∞>[1+2/(2x-1)]^x
=lim<x→∞>{[1+2/(2x-1)]^(2x-1/2)}^[2x/(2x-1)]
=e^lim<x→∞>[2x/(2x-1)]
=e
(3)
lim<x→0>(1+2tanx)^cotx
=lim<x→0>[(1+2tanx)^(1/2tanx)]^(2tanx*cotx)
=e²
lim<n→∞>(n/2)R²sin(2π/n)极限不存在
(3)
lim<x→0>arctan3x/sin2x
=lim<x→0>(arctan3x)'/(sin2x)'
=lim<x→0>[3/(1+9x²)]/(2cos2x)
=[3/(1+0)]/(2×1)
=3/2
(5)
lim<x→0>(1-cos4x)/(xsinx)
=lim<x→0>[1-(1-2sin²2x)]/(xsinx)
=lim<x→0>2sin²2x/(xsinx)
=lim<x→0>2(2sinxcosx)²/(xsinx)
=lim<x→0>8sin²xcos²x/(xsinx)
=lim<x→0>8sinx/x
=8
(1)
lim<x→∞>[x/(1+x)]^(x-3)
=lim<x→∞>[1-(1/1+x)]^(x-3)
=lim<x→∞>{[1+(-1/1+x)]^-(1+x)}^[(x-3)/-(1+x)]
=e^lim<x→∞>[(-x+3)/(x+1)]
=e^(-1)
=1/e
(2)
lim<x→∞>[(2x+1)/(2x-1)]^x
=lim<x→∞>[(2x-1+2)/(2x-1)]^x
=lim<x→∞>[1+2/(2x-1)]^x
=lim<x→∞>{[1+2/(2x-1)]^(2x-1/2)}^[2x/(2x-1)]
=e^lim<x→∞>[2x/(2x-1)]
=e
(3)
lim<x→0>(1+2tanx)^cotx
=lim<x→0>[(1+2tanx)^(1/2tanx)]^(2tanx*cotx)
=e²
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你好样儿的
第一题第一问错了
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