高数—函数极限
limx→0 ((1+3x)开3次根-(1-2x)开3次根)/(x+x²)
=limx→0 [(1+3x)^(1/3)-(1-2x)^(1/3)]/(x+x^2)
0比0型未定式,用罗比达法则。
=limx→0 {(1/3)[(1+3x)^(-2/3)]*(3)-(1/3)[(1-2x)^(-2/3)]*(-2)}/(1+2x)
=limx→0 {(1+3x)^(-2/3)+(2/3)[(1-2x)^(-2/3)]}/(1+2x)
={(1+0)^(-2/3)+(2/3)[(1-0)^(-2/3)]}/(1+0)
={1+(2/3)*1}/1
=5/3
如有不妥之处请回复。谢谢采纳。