Question

请问这道高等数学题怎么做？

步骤要详细点，x不为0

Answer 1

cosx/2cosx/4…cosx/2^n
=cosx/2^ncosx/2^(n-1)…cosx/4cosx/2
=[1/(2sinx/2^n)]2sinx/2^ncosx/2^ncosx/2^(n-1)…cosx/4cosx/2
=[1/(2^2sinx/2^n)]2sinx/2^(n-1)cosx/2^(n-1)…cosx/4cosx/2
…………………………
={1/[2^(n-2)sinx/2^n]}sinx/4cosx/4cosx/2
={1/[2^(n-1)sinx/2^n]}2sinx/4cosx/4cosx/2
={1/[2^(n-1)sinx/2^n]}sinx/2cosx/2
={1/[2^nsinx/2^n]}2sinx/2cosx/2
=sinx/(2^nsinx/2^n)
=xsinx/[(sinx/2^n)/(x/2^n)]

lim(n-->无穷)cosx/2cosx/4…cosx/2^n=lim(n-->无穷)xsinx/[(sinx/2^n)/(x/2^n)]=xsinx