3个回答
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设y=[(1+tanx)/(1+sinx)]^(x^3)
lny=(1/x^3)ln[(1+tanx)/(1+sinx)]
lim[x→0] lny
=lim[x→0] (1/x^3)ln[(1+tanx)/(1+sinx)]
=lim[x→0] (1/x^3)ln[(1+sinx-sinx+tanx)/(1+sinx)]
=lim[x→0] (1/x^3)ln[1+(tanx-sinx)/(1+sinx)]
ln[1+(tanx-sinx)/(1+sinx)]等价于(tanx-sinx)/(1+sinx)
=lim[x→0] (1/x^3)(tanx-sinx)/(1+sinx)
=lim[x→0] (tanx-sinx)/[x^3(1+sinx)]
=lim[x→0] tanx(1-cosx)/[x^3(1+sinx)]
tanx等价于x,1-cosx等价于x^2/2
=lim[x→0] x*(x^2/2)/[x^3(1+sinx)]
=lim[x→0] 1/2(1+sinx)]
=1/2
因此lim[x→0] lny=1/2,则lim[x→0] y=e^(1/2)
看看有没有问题~
lny=(1/x^3)ln[(1+tanx)/(1+sinx)]
lim[x→0] lny
=lim[x→0] (1/x^3)ln[(1+tanx)/(1+sinx)]
=lim[x→0] (1/x^3)ln[(1+sinx-sinx+tanx)/(1+sinx)]
=lim[x→0] (1/x^3)ln[1+(tanx-sinx)/(1+sinx)]
ln[1+(tanx-sinx)/(1+sinx)]等价于(tanx-sinx)/(1+sinx)
=lim[x→0] (1/x^3)(tanx-sinx)/(1+sinx)
=lim[x→0] (tanx-sinx)/[x^3(1+sinx)]
=lim[x→0] tanx(1-cosx)/[x^3(1+sinx)]
tanx等价于x,1-cosx等价于x^2/2
=lim[x→0] x*(x^2/2)/[x^3(1+sinx)]
=lim[x→0] 1/2(1+sinx)]
=1/2
因此lim[x→0] lny=1/2,则lim[x→0] y=e^(1/2)
看看有没有问题~
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