求极限，洛必达法则
1个回答
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x->0
分母
[ (1+(sinx)^6] ^(1/3) - 1
~ ( (1+x^6 ) ^(1/3) - 1
~ 1 + (1/3)x^6 -1
~ (1/3)x^6
lim(x->0) ∫(0-x^3) ln(2t+1) dt /{ [ (1+(sinx)^6] ^(1/3) - 1 ]
=lim(x->0) ∫(0-x^3) ln(2t+1) dt /[ (1/3)x^6 ] (0/0 分子分母分别求导)
=lim(x->0) (3x^2). ln(2x^3+1) /( 2x^5 )
=(3/2)lim(x->0) ln(1+2x^3) / x^3
=(3/2)lim(x->0) (2x^3) / x^3
=3
分母
[ (1+(sinx)^6] ^(1/3) - 1
~ ( (1+x^6 ) ^(1/3) - 1
~ 1 + (1/3)x^6 -1
~ (1/3)x^6
lim(x->0) ∫(0-x^3) ln(2t+1) dt /{ [ (1+(sinx)^6] ^(1/3) - 1 ]
=lim(x->0) ∫(0-x^3) ln(2t+1) dt /[ (1/3)x^6 ] (0/0 分子分母分别求导)
=lim(x->0) (3x^2). ln(2x^3+1) /( 2x^5 )
=(3/2)lim(x->0) ln(1+2x^3) / x^3
=(3/2)lim(x->0) (2x^3) / x^3
=3
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