求极限!!x→0时,[(1+x^4)^
[(1+x^4)^0。5-(1-x^2)^(3/2)]/(4x^2-3x^3)
=[(1+x^4)-(1-x^2)^3]/(4x^2-3x^3)[(1+x^4)^0。5+(1-x^2)^(3/2)]
=(x^6-2x^4+3x^2)/(4x^2-3x^3)[(1+x^4)^0。 5+(1-x^2)^(3/2)]
=(x^4-2x^2+3)/(4-3x)[(1+x^4)^0。5+(1-x^2)^(3/2)]
x→0时,[(1+x^4)^0。5-(1-x^2)^(3/2)]/(4x^2-3x^3)
=(0+0+3)/(4-0)(1+1)=3/8。全部
[(1+x^4)^0。5-(1-x^2)^(3/2)]/(4x^2-3x^3)
=[(1+x^4)-(1-x^2)^3]/(4x^2-3x^3)[(1+x^4)^0。5+(1-x^2)^(3/2)]
=(x^6-2x^4+3x^2)/(4x^2-3x^3)[(1+x^4)^0。
5+(1-x^2)^(3/2)]
=(x^4-2x^2+3)/(4-3x)[(1+x^4)^0。5+(1-x^2)^(3/2)]
x→0时,[(1+x^4)^0。5-(1-x^2)^(3/2)]/(4x^2-3x^3)
=(0+0+3)/(4-0)(1+1)=3/8。收起