2^n.sinπ/2^n (n趋近无穷),求极限,用两个重要极限公式求_数学解答

2^n.sinπ/2^n (n趋近无穷),求极限,用两个重要极限公式求

人气：133 ℃　时间：2020-05-23 15:22:29

解答

=lim[sin(π/2^n)/(π/2^n)]·n
=1·n
=n=lim[sin(π/2^n)/(π/2^n)]·n
=1·π
=π
按错了符号=lim[sin(π/2^n)/(π/2^n)]·π
=1·π
=π