1个回答
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(2)
(2x+3)/(2x+1) = 1 + 2/(2x+1)
let
1/y = 2/(2x+1)
lim(x->∞) [ (2x+3)/(2x+1) ]^(x+1)
=lim(x->∞) [ 1+ 2/(2x+1) ]^(x+1)
=lim(y->∞) [ 1+ 1/y ]^( (2y-1)/2+1)
=lim(y->∞) [ 1+ 1/y ]^y
=e
(4)
lim(x->1+) [(x^2-1)/(x-1) ].e^[-1/(x-1) ]
=lim(x->1+) (x+1) .e^[-1/(x-1) ]
=0
(6)
lim(x->0) ln(sinx/x)
=ln1
=0
(8)
lim(x->∞) { x +[x+x^(1/2)]^(1/2) } ^(1/2) - x^(1/2)
=lim(x->∞) [x+x^(1/2)]^(1/2) / [ { x +[x+x^(1/2)]^(1/2) } ^(1/2) + x^(1/2) ]
=lim(x->∞) [1+x^(-1/2)]^(1/2) / [ { 1 +[x^(-1)+x^(-3/2)]^(1/2) } ^(1/2) + 1 ]
=1/(1+1)
=1/2
(2x+3)/(2x+1) = 1 + 2/(2x+1)
let
1/y = 2/(2x+1)
lim(x->∞) [ (2x+3)/(2x+1) ]^(x+1)
=lim(x->∞) [ 1+ 2/(2x+1) ]^(x+1)
=lim(y->∞) [ 1+ 1/y ]^( (2y-1)/2+1)
=lim(y->∞) [ 1+ 1/y ]^y
=e
(4)
lim(x->1+) [(x^2-1)/(x-1) ].e^[-1/(x-1) ]
=lim(x->1+) (x+1) .e^[-1/(x-1) ]
=0
(6)
lim(x->0) ln(sinx/x)
=ln1
=0
(8)
lim(x->∞) { x +[x+x^(1/2)]^(1/2) } ^(1/2) - x^(1/2)
=lim(x->∞) [x+x^(1/2)]^(1/2) / [ { x +[x+x^(1/2)]^(1/2) } ^(1/2) + x^(1/2) ]
=lim(x->∞) [1+x^(-1/2)]^(1/2) / [ { 1 +[x^(-1)+x^(-3/2)]^(1/2) } ^(1/2) + 1 ]
=1/(1+1)
=1/2
追问
^是什么意思啊
亲亲
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