求lim(x→0)(1-con2x)/xSinx的极限,
问题描述:
求lim(x→0)(1-con2x)/xSinx的极限,
答
原式=lim(x->0)[2sin²x/(xsinx)] (应用三角函数倍角公式)
=lim(x->0)(2sinx/x)
=2*lim(x->0)(sinx/x)
=2*1 (应用重要极限lim(x->0)(sinx/x)=1)
=2