求2^nsinx/2^n在n趋向于无穷的时候的极限

hwdy_19832022-10-04 11:39:541条回答

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mafu_wang 共回答了21个问题 | 采纳率90.5%
=limxsin(x/2^n)/(x/2^n)=x
1年前

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LOVE19861年前1
lewyoung 共回答了19个问题 | 采纳率94.7%
①等价无穷小量替换:
lim(n→∞)2^nsin(x/2^n)
=lim(n→∞)2^n*(x/2^n)
= x

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【罗必塔法则】
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求极限:lim(n→∞)2^nsinx/2^n(x不为零的常数);
鬼遁猫妖1年前2
shadowfly88 共回答了16个问题 | 采纳率81.3%
等价无穷小量替换:
lim(n→∞)2^nsin(x/2^n)
=lim(n→∞)2^n*(x/2^n)
= x