请问这道求极限怎么做呀?
2个回答
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consider
lim(x->∞){ [ 1^(1/x) +2^(1/x)+...+2017^(1/x)]/2017 } ^x
let
y=1/x
L=lim(x->∞){ [ 1^(1/x) +2^(1/x)+...+2017^(1/x)]/2017 } ^x
=lim(y->0){ [ 1^y +2^y+...+2017^y]/2017 } ^(1/y)
lnL
=lim(y->0) ln{ [ 1^y +2^y+...+2017^y]/2017 } /y (0/0)
=lim(y->0) [(ln1).1^y+(ln2).2^y+...+(ln2017).2017^y ]/ [ 1^y +2^y+...+2017^y]
=lim(y->0) [(ln1).(1/2017)^y+(ln2).(2/2017)^y+...+(ln2017) ]
/ [ (1/2017)^y +(2/2017)^y+...+1]
=ln2017
=>
L = 2017
=>
lim(n->∞){ [ 1^(1/n) +2^(1/n)+...+2017^(1/n)]/2017 } ^n =2017
lim(x->∞){ [ 1^(1/x) +2^(1/x)+...+2017^(1/x)]/2017 } ^x
let
y=1/x
L=lim(x->∞){ [ 1^(1/x) +2^(1/x)+...+2017^(1/x)]/2017 } ^x
=lim(y->0){ [ 1^y +2^y+...+2017^y]/2017 } ^(1/y)
lnL
=lim(y->0) ln{ [ 1^y +2^y+...+2017^y]/2017 } /y (0/0)
=lim(y->0) [(ln1).1^y+(ln2).2^y+...+(ln2017).2017^y ]/ [ 1^y +2^y+...+2017^y]
=lim(y->0) [(ln1).(1/2017)^y+(ln2).(2/2017)^y+...+(ln2017) ]
/ [ (1/2017)^y +(2/2017)^y+...+1]
=ln2017
=>
L = 2017
=>
lim(n->∞){ [ 1^(1/n) +2^(1/n)+...+2017^(1/n)]/2017 } ^n =2017
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