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令t=1/x
则左边=lim(t->0) [(1+t)^(1/t)-e]/t
=lim(t->0) {e^[(1/t)*ln(1+t)]-e}/t
=lim(t->0) e*{e^[(1/t)*ln(1+t)-1]-1}/t
=lim(t->0) e*[(1/t)*ln(1+t)-1]/t
=e*lim(t->0) [ln(1+t)-t]/t^2
=e*lim(t->0) [1/(1+t)-1]/2t
=e*lim(t->0) (-t)/2t(1+t)
=e*lim(t->0) -1/2(1+t)
=-e/2
=右边
则左边=lim(t->0) [(1+t)^(1/t)-e]/t
=lim(t->0) {e^[(1/t)*ln(1+t)]-e}/t
=lim(t->0) e*{e^[(1/t)*ln(1+t)-1]-1}/t
=lim(t->0) e*[(1/t)*ln(1+t)-1]/t
=e*lim(t->0) [ln(1+t)-t]/t^2
=e*lim(t->0) [1/(1+t)-1]/2t
=e*lim(t->0) (-t)/2t(1+t)
=e*lim(t->0) -1/2(1+t)
=-e/2
=右边
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用了洛必达吗?
刚大一有点看不懂(#-.-)
展开全部
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) ]
=e.e^[-(1/2)y +o(y) ]
lim(x->+∞) x[( 1+ 1/x)^x - e]
y=1/x
=lim(y->0+) [( 1+ y)^(1/y) - e] /y
=lim(y->0+) [e.e^[-(1/2)y ] - e] /y
=lim(y->0+) e.{ e^[-(1/2)y ] - 1} /y
=lim(y->0+) e.[ -(1/2)y] /y
=-(1/2)e
( 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) ]
=e.e^[-(1/2)y +o(y) ]
lim(x->+∞) x[( 1+ 1/x)^x - e]
y=1/x
=lim(y->0+) [( 1+ y)^(1/y) - e] /y
=lim(y->0+) [e.e^[-(1/2)y ] - e] /y
=lim(y->0+) e.{ e^[-(1/2)y ] - 1} /y
=lim(y->0+) e.[ -(1/2)y] /y
=-(1/2)e
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