limx→1 {[x^(1/3)]-1}/{[x^(1/2)]-1}的极限多少
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limx→1 {[x^(1/3)]-1}/{[x^(1/2)]-1}的极限多少
![limx→1 {[x^(1/3)]-1}/{[x^(1/2)]-1}的极限多少](/uploads/image/z/17337061-37-1.jpg?t=limx%E2%86%921+%7B%5Bx%5E%281%2F3%29%5D-1%7D%2F%7B%5Bx%5E%281%2F2%29%5D-1%7D%E7%9A%84%E6%9E%81%E9%99%90%E5%A4%9A%E5%B0%91)
令 x=t^6 ,由 x 趋于 1 得 t 趋于 1 ,
原式 = lim(t→1) [(t^2-1)/(t^3-1)]
= lim(t→1) [(t+1)(t-1)] / [(t-1)(t^2+t+1)]
= lim(t→1) (t+1)/(t^2+t+1)
= (1+1)/(1+1+1)
= 2/3
原式 = lim(t→1) [(t^2-1)/(t^3-1)]
= lim(t→1) [(t+1)(t-1)] / [(t-1)(t^2+t+1)]
= lim(t→1) (t+1)/(t^2+t+1)
= (1+1)/(1+1+1)
= 2/3
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