> 数学 >
2^n*sin(x/2^n),n→∞的极限(x≠0)
人气:303 ℃ 时间:2020-09-30 05:54:47
解答
lim 2^n*sin(x/2^n)
=lim sin(x/2^n)/(1/2^n) (sin(x/2^n)用其等价无穷小1/2^n替代)
=lim (x/2^n)/(1/2^n)
=x
推荐
猜你喜欢
© 2026 79432.Com All Rights Reserved.
电脑版|手机版