Question

高数。求极限。正确的做法是什么？过程。。。。

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Answer 1

ln(3/5)<ln1=0，所以是e^(-∞)=0

追答

第二个一样，要注意正负

分子有理化没错，然后利用无穷大倒数为零分子分母同除以，=lim(1+0)/(1+1)

Answer 2

lim(x->∞) (3/5)^x =0
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L=lim(x->∞) [x/(3x+1)]^x
lnL
=lim(x->∞) ln[x/(3x+1)] / (1/x) (0/0)
=lim(x->∞) [1/x -3/(3x+1)] / (-1/x^2)
=lim(x->∞) -x^2. [(3x+1)-3x ]/[x(3x+1)]
=lim(x->∞) -x^2/[x(3x+1)]
=lim(x->∞) -1/(3+1/x)
=-1/3
=>

lim(x->∞) [x/(3x+1)]^x = e^(-1/3)
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lim(x->∞) [ √(x^2+x) - √(x^2-1) ]
=lim(x->∞) (x+1)/[ √(x^2+x) + √(x^2-1) ]
=lim(x->∞) (1+1/x)/[ √(1+1/x) + √(1-1/x) ]
= 1/(1+1)
=1/2