高数求极限,这两题怎么做呢? 20
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(5)
lim(n->∞) √n. [ (n^2-1)^(1/4) -√(n+1) ]
u=1/x
x->0+
分子
(1-x^2)^(1/4) = 1-(1/4)x^2+o(x^2)
√(1+x) = 1+(1/2)x +o(x)
(1-x^2)^(1/4) -√(1+x) = -(1/2)x +o(x)
//
consider
lim(u->∞) √u. [ (u^2-1)^(1/4) -√(u+1) ]
=lim(x->0+) √(1/x). [ (1/x^2 -1)^(1/4) -√(1/x+1) ]
=lim(x->0+) [ (1-x^2)^(1/4) -√(1+x) ] /x
=lim(x->0+) (-1/2)x /x
=-1/2
=>
lim(n->∞) √n. [ (n^2-1)^(1/4) -√(n+1) ] =-1/2
(2)
分子分母同时除(n^3)
lim(n->∞) (n^3+2n^2-3n+1)/(n^2+1)
=lim(n->∞) (1+2/n-3/n^2+1/n^3)/(1/n+1/n^3)
->∞
分子->1
分母->0
lim(n->∞) √n. [ (n^2-1)^(1/4) -√(n+1) ]
u=1/x
x->0+
分子
(1-x^2)^(1/4) = 1-(1/4)x^2+o(x^2)
√(1+x) = 1+(1/2)x +o(x)
(1-x^2)^(1/4) -√(1+x) = -(1/2)x +o(x)
//
consider
lim(u->∞) √u. [ (u^2-1)^(1/4) -√(u+1) ]
=lim(x->0+) √(1/x). [ (1/x^2 -1)^(1/4) -√(1/x+1) ]
=lim(x->0+) [ (1-x^2)^(1/4) -√(1+x) ] /x
=lim(x->0+) (-1/2)x /x
=-1/2
=>
lim(n->∞) √n. [ (n^2-1)^(1/4) -√(n+1) ] =-1/2
(2)
分子分母同时除(n^3)
lim(n->∞) (n^3+2n^2-3n+1)/(n^2+1)
=lim(n->∞) (1+2/n-3/n^2+1/n^3)/(1/n+1/n^3)
->∞
分子->1
分母->0
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