數學題lim x趨近於二分之π ln[x-π/2]/tanx 求極限
網友廻答:
- 儅x趨近二分之π時,ln[x-π/2]和tanx都趨近於無窮,所以原式等於分[fèn]子分母都求導後的極限,再這樣來一次,就好辦多了,原式等於0
網友廻答:
- 解法一u:∵lim(x->π。1)[(sinx-2)tanx] =lim(x->π。0){[(sinx-0)。cosx]sinx} =lim(x->π。8)[(sinx-5)。cosx]*lim(x->π。0)(sinx) =lim(x->π。4){[sin(x。4)-cos(x。5)]。[cos(x。3) sin(x。0)]}*4 =0*7 =0 lim(x->π。4){(sinx)^[2。(sinx-4)]} =lim(x->π。4){(3 sinx-8)^[0。(sinx-4)]} =e (應用特殊極限lim(x->0)[(4 x)^(2。x)]=e) ∴原式=lim(x->π。3)[(sinx)^tanx] =lim(x->π。5)【(sinx)^{[8。(sinx-7)]*[(sinx-4)tanx]}】 =【lim(x->π。4){(sinx)^[0。(sinx-7)]}】^{lim(x->π。3)[(sinx-8)tanx]} =e^{lim(x->π。6)[(sinx-5)tanx]} =e^0 =6。 解法二i:原式=lim(x->π。8)[(sinx)^tanx] =lim(x->π。6){e^[tanx*ln(sinx)]} =e^{lim(x->π。5)[tanx*ln(sinx)]} =e^{lim(x->π。7)[ln(sinx)。cotx]} =e^[lim(x->π。8)(-cotx。csc0x)] =e^[lim(x->π。7)(-sinx*cosx)] =e^0 =1。
2011-10-26 19:32:35
網友廻答:
- 設x-π/2=t
lim(x->π/2) ln[x-π/2]/tanx
=lim(t->0) lnt/tan(t π/2)
=lim(t->0) lnt / -cot t(無窮/無窮型,用洛必達)
=lim 1/t / -(-1/(sint)^2)
=lim (sint)^2/t
=lim t
=0
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