用洛比达法则求极限?1 lim sinx-sina /(x^2-a^2) (a≠0)x→a2 lim (cosax-cosbx)/x^2 (a,b≠0)x→0

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/04 14:59:45

用洛比达法则求极限?1 lim sinx-sina /(x^2-a^2) (a≠0)x→a2 lim (cosax-cosbx)/x^2 (a,b≠0)x→0
用洛比达法则求极限?
1 lim sinx-sina /(x^2-a^2) (a≠0)
x→a
2 lim (cosax-cosbx)/x^2 (a,b≠0)
x→0

用洛比达法则求极限?1 lim sinx-sina /(x^2-a^2) (a≠0)x→a2 lim (cosax-cosbx)/x^2 (a,b≠0)x→0
1.lim sinx-sina /(x^2-a^2)=lim cosx/2x=cosa/2a
2.lim (cosax-cosbx)/x^2 =lim (-a*sinax+b*sinbx)/2x
=lim [-(a^2) *cosax+(b^2) *cosbx]/2=(b^2-a^2)/2