极限的计算问题
∫sin(xt)dt
=(1/x)∫sin(xt)d(xt)
令xt=u,则:t=x^2时,u=x^3;t=x时,u=x^2
=(1/x)∫sinudu
所以,原极限=lim(1/x)∫sinudu/x^2
=lim∫sinudu/x^3
=lim[sin(x^3)-sin(x^2)]/(3x^2)
=(1/3)*lim[cos(x^3)*3x^2-cos(x^2)*2x]/(2x)
=(1/3)*lim[cos(x^3)*3x-2*cos(x^2)]/2
=(1/3)*(0-2)/2
=-1/3。
∫sin(xt)dt
=(1/x)∫sin(xt)d(xt)
令xt=u,则:t=x^2时,u=x^3;t=x时,u=x^2
=(1/x)∫sinudu
所以,原极限=lim(1/x)∫sinudu/x^2
=lim∫sinudu/x^3
=lim[sin(x^3)-sin(x^2)]/(3x^2)
=(1/3)*lim[cos(x^3)*3x^2-cos(x^2)*2x]/(2x)
=(1/3)*lim[cos(x^3)*3x-2*cos(x^2)]/2
=(1/3)*(0-2)/2
=-1/3。
收起