lim(x→0)[(sinx)^2-x^2(cosx)^2]/[x(e^2x-1)ln[1+(tanx)^2]]
问题描述:
lim(x→0)[(sinx)^2-x^2(cosx)^2]/[x(e^2x-1)ln[1+(tanx)^2]]
答
lim(x→0)[(sinx)^2-x^2(cosx)^2]/[x(e^2x-1)ln[1+(tanx)^2]]
=lim(x→0)[(sinx)^2-x^2(cosx)^2]/[x*2x*(tanx)^2]
=lim(x→0)[(sinx)^2-x^2(cosx)^2]/[x*2x*x^2]
=lim(x→0)[(sinx)^2-x^2(cosx)^2]/[2x^4](0/0)
=lim(x→0)[2sinxcosx-2x(cosx)^2+2x^2sinxcosx]/(8x^3)
=lim(x→0)cosx[2sinx-2x(cosx)+2x^2sinx]/(8x^3)
=lim(x→0)[2sinx-2x(cosx)+2x^2sinx]/(8x^3)
=lim(x→0)(2sinx-2xcosx+2x^2sinx)/(8x^3)(0/0)
=lim(x→0)(2cosx-2cosx+2xsinx+4xsinx+2x^2cosx)/(24x^2)
=lim(x→0)(6xsinx+2x^2cosx)/(24x^2)
=lim(x→0)2x(3sinx+xcosx)/(24x^2)
=lim(x→0)(3sinx+xcosx)/(12x) (0/0)
=lim(x→0)(3cosx+cosx-xsinx)/12
=4/12
=1/3