lim(1-cosx)^1/lnx

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lim(1-cosx)^1/lnx
lim【x→0+】(1-cosx)^(1/lnx)

lim【x→0+】(1-cosx)^(1/lnx)lim【x→0+】(1-cosx)^(1/lnx)lim【x→0+】(1-cosx)^(1/lnx)y=(1-cosx)^(1/lnx)lny=(1/

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lim(lnx)-1/(x-e)

lim(lnx)-1/(x-e)lim(lnx)-1/(x-e)lim(lnx)-1/(x-e)lim(x→e)(lnx-1)/(x-e)=lim(x→e)(1/x)/1=lim(x→e)1/x=1/

1/x/sinx/1-cosx=1-cosx/xsinx=1/2 这个是怎么变的啊原题是lnx/ln(1-cosx) lim趋于0+

1/x/sinx/1-cosx=1-cosx/xsinx=1/2这个是怎么变的啊原题是lnx/ln(1-cosx)lim趋于0+1/x/sinx/1-cosx=1-cosx/xsinx=1/2这个是怎

lim(tan5x-cosx+1)/sin3x

lim(tan5x-cosx+1)/sin3xlim(tan5x-cosx+1)/sin3xlim(tan5x-cosx+1)/sin3xlim(x-->0)(tan5x-cosx+1)/sin3x=

lim(x趋于无穷) (2lnx+sinx)/(lnx+cosx)rt

lim(x趋于无穷)(2lnx+sinx)/(lnx+cosx)rtlim(x趋于无穷)(2lnx+sinx)/(lnx+cosx)rtlim(x趋于无穷)(2lnx+sinx)/(lnx+cosx)

lim 1(x/(x-1)-1/(lnx))

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lim(x→1)((1-x)/lnx)lim(x→1)((1-x)/lnx)lim(x→1)((1-x)/lnx)-1

lim→0+ lnx ln(1+X)

lim→0+lnxln(1+X)lim→0+lnxln(1+X)lim→0+lnxln(1+X)lim→0+lnxln(1+X)=lim→0+ln(1+X)/(1/inx)运用洛比达法则=lim→0+

为什么lim sinx/(x-π) =lim cosx/1

为什么limsinx/(x-π)=limcosx/1为什么limsinx/(x-π)=limcosx/1为什么limsinx/(x-π)=limcosx/1你这个是x趋于π/2吧,这是一个0/0的类型

急、一道简单的数学题、、、极限为什么这题lim(x→0+)(1-cosx)^[1/(1+lnx)]用等价无穷小和用洛必达法则做出来的答案不同?看我做一次,等价无穷小的,原式=exp{lim(x→0+)[ln(1-cosx)]/(1+lnx)}=exp{lim(x→0+)-

急、一道简单的数学题、、、极限为什么这题lim(x→0+)(1-cosx)^[1/(1+lnx)]用等价无穷小和用洛必达法则做出来的答案不同?看我做一次,等价无穷小的,原式=exp{lim(x→0+)

lim(1-cosx)x趋向0,

lim(1-cosx)x趋向0,lim(1-cosx)x趋向0,lim(1-cosx)x趋向0,lim(1-cosx)x趋向0=1-cos0°=1-1=0答案是0当x趋向0时,cosx趋向1,1-1=

lim(x—0) (1-cosx)/x

lim(x—0)(1-cosx)/xlim(x—0)(1-cosx)/x lim(x—0)(1-cosx)/x不是1+sin(x),是sin(x).再对它求极限,可得0

lim(x→0)(1/cosx)=?

lim(x→0)(1/cosx)=?lim(x→0)(1/cosx)=?lim(x→0)(1/cosx)=?1/cosx在x=0处连续,直接代值即可lim(x→0)(1/cosx)=1/cos0=1l

lim(cosx)^(1/(1-cosx)).x趋向于0

lim(cosx)^(1/(1-cosx)).x趋向于0lim(cosx)^(1/(1-cosx)).x趋向于0lim(cosx)^(1/(1-cosx)).x趋向于0原式=lim(x->0){[1+

大数lim(cosx)^(1/(1-cosx)).x趋向于0

大数lim(cosx)^(1/(1-cosx)).x趋向于0大数lim(cosx)^(1/(1-cosx)).x趋向于0大数lim(cosx)^(1/(1-cosx)).x趋向于0原式=lim(x→0

求Lim(x→0)(sinx/x)^(cosx/1-cosx)

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lim(x趋于1)1/lnx-1/x-1=

lim(x趋于1)1/lnx-1/x-1=lim(x趋于1)1/lnx-1/x-1=lim(x趋于1)1/lnx-1/x-1=采纳后告诉你,

lim(x/x-1-1/lnx)x趋于1

lim(x/x-1-1/lnx)x趋于1lim(x/x-1-1/lnx)x趋于1lim(x/x-1-1/lnx)x趋于1先通分,得lim(xlnx-x+1)/(lnx(x-1))把x=1代进去,分子分

lim(x趋向于1)[x/(1-x)-1/lnx]

lim(x趋向于1)[x/(1-x)-1/lnx]lim(x趋向于1)[x/(1-x)-1/lnx]lim(x趋向于1)[x/(1-x)-1/lnx]原式=lim(x->1)[xlnx-(1-x)]/