超一流数学高手、数学教师请进(极限027)

设y=1/x,则当x->∞时，y->0
∴原式=(y->0)lim[(siny+cosy)^(1/y)]
=e^{(y->0)lim[ln(siny+cosy)/y]}
=e^{(y->0)lim[(cosy-siny)/(siny+cosy)]} (用一次罗比达法)
=e^1=e
另一解法：
设y=1/x,则当x->∞时，y->0.
∵(siny+cosy)^(1/y)=(1+siny/cosy)^(1/y)*(cosy)^(1/y)
=[(1+siny/cosy)^(cosy/siny)]^[siny/(ycosy)]*
*[(1+cosy-1)^(1/cosy-1)]^[(cosy-1)/y]
又 (y->0)lim[(1+siny/cosy)^(cosy/siny)]=e
(y->0)lim[siny/(ycosy)]=(y->0)lim[(siny/y)*(1/cosy)]=1*1=1
(y->0)lim[(1+cosy-1)^(1/cosy-1)]=e
(y->0)lim[(cosy-1)/y]=(y->0)lim{[sin(y/2)/(y/2)]²*（-y/2)=1*0=0
∴原式=(e^1)*(e^0)=e.

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