当x趋近于0时lim[x平方/2+1-根号下(1+x平方)]/[(cosx-e的x^2次幂)ln(1-sinx^2)]=?

Question

当x趋近于0时lim[x平方/2+1-根号下(1+x平方)]/[(cosx-e的x^2次幂)ln(1-sinx^2)]=?

Answer 1

lim(x→0) [x^2/2+1-√(1+x^2)]/[(cosx-e^x^2)ln(1-sinx^2)]
=lim(x→0) [x^2/2+1-√(1+x^2)]/[(cosx-e^x^2)(-sinx^2)]
=lim(x→0) [x^2/2+1-√(1+x^2)]/[(cosx-e^x^2)(-x^2)] (0/0)
=lim(x→0) [x-x/√(1+x^2)]/[(-sinx-2xe^x^2)(-x^2)-2x(cosx-e^x^2)]
=lim(x→0) [√(1+x^2)-1]/[(-sinx-2xe^x^2)(-x)-2(cosx-e^x^2)]
=lim(x→0) (x^2/2)/(xsinx+2x^2e^x^2+2cosx-2e^x^2) (0/0)
=lim(x→0) x/(sinx+xcosx+4xe^x^2+4x^4e^x^2-2sinx-4xe^x^2)
=lim(x→0) x/(xcosx+4xe^x^2-sinx) (0/0)
=lim(x→0) 1/(cosx-xsinx+4e^x^2+8x^2e^x^2-cosx)
=lim(x→0) 1/(-xsinx+4e^x^2+8x^2e^x^2)
=1/4