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约定:lim[n→∞]表示求n趋于无穷大时的极限,lim[x→0]表示求x趋0时的极限。
x≠0,设u=x/2^n
cos(x/2)·cos(x/2²)·...·cos(x/2^n)
=(2^n·sin(x/2^n)·cos(x/2)·cos(x/2²)····cos(x/2^n))/(2^n·sin(x/2^n))
=(sinx)/(2^n·sin(x/2^n))
=(sinx)/(x·(sinu/u))
左边=lim[x→0](lim[n→∞]((sinx)/(x·(sinu/u))))
=lim[x→0]((sinx)/x)·(lim[n→∞](1/(sinu/u))
=lim[x→0]((sinx)/x)·(1/1)
=lim[x→0]((sinx)/x)
=1
所以 lim[x→0](lim[n→∞](cos(x/2)·cos(x/2²)·...·cos(x/2^n))=1
x≠0,设u=x/2^n
cos(x/2)·cos(x/2²)·...·cos(x/2^n)
=(2^n·sin(x/2^n)·cos(x/2)·cos(x/2²)····cos(x/2^n))/(2^n·sin(x/2^n))
=(sinx)/(2^n·sin(x/2^n))
=(sinx)/(x·(sinu/u))
左边=lim[x→0](lim[n→∞]((sinx)/(x·(sinu/u))))
=lim[x→0]((sinx)/x)·(lim[n→∞](1/(sinu/u))
=lim[x→0]((sinx)/x)·(1/1)
=lim[x→0]((sinx)/x)
=1
所以 lim[x→0](lim[n→∞](cos(x/2)·cos(x/2²)·...·cos(x/2^n))=1
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