用数列极限定义证明lim√narctann/1+n=0
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|√n. (arctann) /(1+n) -0 |<ε
(π/2) [√n/(1+n) ]<ε
(π/2)/√n < ε
n> (π/(2ε))^2
Choose N =[(π/(2ε))^2] +1
ie
∀ε>0 ,∃N =[(π/(2ε))^2] +1 st
|√n. (arctann) /(1+n) -0 |<ε , ∀n>N
=>
lim(n->∞) [√n. (arctann) /(1+n) ] =0
(π/2) [√n/(1+n) ]<ε
(π/2)/√n < ε
n> (π/(2ε))^2
Choose N =[(π/(2ε))^2] +1
ie
∀ε>0 ,∃N =[(π/(2ε))^2] +1 st
|√n. (arctann) /(1+n) -0 |<ε , ∀n>N
=>
lim(n->∞) [√n. (arctann) /(1+n) ] =0
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