求n趋近于无穷大时的极限limcos(φ/2)cos(φ/2^2).cos(φ/2^n),

首先把2倍角sin2x=2sinxcosx推广下，有:sin[φ/2^(n-1)]=2sin(φ/2^n)cos(φ/2^n)，然后解题。
1)当φ=0是，原式=lim(cos0*cos0*...)=lim1=1
2)当φ≠0时，有：
cos(φ/2)cos(φ/2^2)........cos(φ/2^n)
=[cos(φ/2)cos(φ/2^2)........cos(φ/2^n)*2^n*sin(φ/2^n)]/[2^n*sin(φ/2^n)]
=sinφ/[2^n*sin(φ/2^n)]
由于sin(φ/2^n)是n->∞的无穷小量，有sin(φ/2^n)~φ/2^n。故：
limcos(φ/2)cos(φ/2^2)........cos(φ/2^n)=lim{sinφ/[2^n*sin(φ/2^n)}=lim(sinφ/[2^n*(φ/2^n)]
=lim(sinφ/φ)
=sinφ/φ

cos(φ/2)cos(φ/2^2)........cos(φ/2^n)=cos(φ/2)cos(φ/2^2)........cos(φ/2^n)sin(φ/2^n)/sin(φ/2^n)
=(1/2)^n *cosφ /sin(φ/2^n)
=cosφ/φ *(φ/2^n) /sin(φ/2^n)
后者极限为1
所以原式极限为 cosφ/φ （φ≠0）
1（φ=0）

φ=0时,原式=1
φ≠0时,原式=imcos(φ/2)cos(φ/2^2).cos(φ/2^n)
=limcos(φ/2)cos(φ/2^2).cos[φ/2^(n-1)]sin[φ/2^(n-1)]/[2sin(φ/2^n)]
=limcos(φ/2)cos(φ/2^2).2cos[φ/2^(n-1)]sin[φ/2^(n-1)]/[(2^2)sin(φ/2^n)]
=limcos(φ/2)cos(φ/2^2).cos[φ/2^(n-2)]sin[φ/2^(n-2)]/[(2^2)sin(φ/2^n)]
=limcos(φ/2)cos(φ/2^2).2cos[φ/2^(n-2)]sin[φ/2^(n-2)]/[(2^3)sin(φ/2^n)]
=limcos(φ/2)cos(φ/2^2).cos[φ/2^(n-3)]sin[φ/2^(n-3)]/[(2^3)sin(φ/2^n)]
=.
=limcos(φ/2)sin(φ/2)/[(2^(n-1))sin(φ/2^n)]
=lim2cos(φ/2)sin(φ/2)/[(2^n)sin(φ/2^n)]
=limsinφ/[(2^n)sin(φ/2^n)]
=lim[sinφ/φ]/[sin(φ/2^n)/(φ/2^n)]
=sinφ/φ