高数极限运算limx(x→∞)2^n*Sin丌⼀2^n

2025-05-24 05:17:12
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回答(1):

lim(n->∞)2^nsin(x/2^n)
设 t=1/2^n,t->0
lim(t->0) sin(xt)/t
=lim(t->0) xsin(xt)/xt
=x*1=x