一道大学数学求极限题目
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一道大学数学求极限题目一道大学数学求极限题目 一道大学数学求极限题目cos(x/2)cos(x/4)...cos(x/2^(n-1))cos(x/2^n)=cos(x/2)cos(x/4).
一道大学数学求极限题目
一道大学数学求极限题目
一道大学数学求极限题目
cos(x/2)cos(x/4)...cos(x/2^(n-1))cos(x/2^n)
= cos(x/2)cos(x/4)...cos(x/2^(n-1))cos(x/2^n)sin(x/2^n)/sin(x/2^n)
= cos(x/2)cos(x/4)...cos(x/2^(n-1))sin(x/2^(n-1))/(2sin(x/2^n))
= cos(x/2)cos(x/4)...sin(x/2^(n-2))/(4sin(x/2^n))
...
= cos(x/2)sin(x/2)/(2^(n-1)·sin(x/2^n))
= sin(x)/(2^n·sin(x/2^n))
= sin(x)/(x·sin(x/2^n)/(x/2^n)).
n → ∞时, x/2^n → 0, 从而sin(x/2^n)/(x/2^n) → 1.
因此所求极限为sin(x)/x.