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