求极限第三题怎么做呢
4个回答
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y->0
( 1+y)^(1/y)
=e^[ln(1+y)/y ]
=e^{ [ y-(1/2)y^2 +o(y^2)] /y }
=e^[1-(1/2)y +o(y)]
( 1+y)^(1/y)/ e = e^[-(1/2)y +o(y)]
lim(x->+∞) [( 1+1/x)^x/e ]^x
y=1/x
lim(y->0) [( 1+y)^(1/y)/e ]^(1/y)
=lim(y->0) { e^[-(1/2)y ] }^(1/y)
=e^(-1/2)
( 1+y)^(1/y)
=e^[ln(1+y)/y ]
=e^{ [ y-(1/2)y^2 +o(y^2)] /y }
=e^[1-(1/2)y +o(y)]
( 1+y)^(1/y)/ e = e^[-(1/2)y +o(y)]
lim(x->+∞) [( 1+1/x)^x/e ]^x
y=1/x
lim(y->0) [( 1+y)^(1/y)/e ]^(1/y)
=lim(y->0) { e^[-(1/2)y ] }^(1/y)
=e^(-1/2)
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展开全部
=e^limxln(1+1/x)^x/e
=e^limx(xln(1+1/x)-1)
t=1/x趋于零
=e^lim(ln(1+t)/t-1)/t
=e^lim(ln(1+t)-t)/t²
=e^lim(1/(1+t)-1)/2t
=e^lim-1/2(1+t)
=e^(-1/2)
=1/√e
=e^limx(xln(1+1/x)-1)
t=1/x趋于零
=e^lim(ln(1+t)/t-1)/t
=e^lim(ln(1+t)-t)/t²
=e^lim(1/(1+t)-1)/2t
=e^lim-1/2(1+t)
=e^(-1/2)
=1/√e
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