大一微积分题求解
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x^(1/3) -1 = [x^(1/6)]^2 -1^2 =[ x^(1/6) -1 ] .[ x^(1/6) +1 ]
x^(1/2) -1 = [x^(1/6)]^3 -1^3 =[ x^(1/6) -1 ] .[ x^(1/3) +x^(1/6) +1 ]
lim(x->1) [x^(1/3) -1]/[x^(1/2) -1]
=lim(x->1) [ x^(1/6) -1 ] .[ x^(1/6) +1 ]/{ [ x^(1/6) -1 ] .[ x^(1/3) +x^(1/6) +1 ] }
=lim(x->1) [ x^(1/6) +1 ]/[ x^(1/3) +x^(1/6) +1 ]
= 2/3
x^(1/2) -1 = [x^(1/6)]^3 -1^3 =[ x^(1/6) -1 ] .[ x^(1/3) +x^(1/6) +1 ]
lim(x->1) [x^(1/3) -1]/[x^(1/2) -1]
=lim(x->1) [ x^(1/6) -1 ] .[ x^(1/6) +1 ]/{ [ x^(1/6) -1 ] .[ x^(1/3) +x^(1/6) +1 ] }
=lim(x->1) [ x^(1/6) +1 ]/[ x^(1/3) +x^(1/6) +1 ]
= 2/3
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详细过程如图所示，希望能帮到你
追答
拍个清楚的，望采纳
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