求极限 lim→无穷大[(x+1)/(x-1)]^x-1 lim→正无穷[√(x^2+x)-√(x^2-x)]
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求极限 lim→无穷大[(x+1)/(x-1)]^x-1 lim→正无穷[√(x^2+x)-√(x^2-x)]
求详解
求详解
1.此为1^∞型,联想到(1+1/x)^(x)~e,x→∞
lim [(x+1)/(x-1)]^(x-1)
x→∞
=lim [1+2/(x-1)]^(x-1)
x→∞
=lim [1+2/(x-1)]^{[(x-1)/2]·2}
x→∞
=e²
2.此题需分子有理化:
lim √(x²+x)-√(x²-x)
x→+∞
=lim [√(x²+x)-√(x²-x)]·[√(x²+x)+√(x²-x)] /[√(x²+x)+√(x²-x)]
x→+∞
=lim 2x/[√(x²+x)+√(x²-x)]
x→+∞
=lim 2/[√(1+1/x)+√(1-1/x)]
x→+∞
=lim 2/(1+1)
x→+∞
=1
lim [(x+1)/(x-1)]^(x-1)
x→∞
=lim [1+2/(x-1)]^(x-1)
x→∞
=lim [1+2/(x-1)]^{[(x-1)/2]·2}
x→∞
=e²
2.此题需分子有理化:
lim √(x²+x)-√(x²-x)
x→+∞
=lim [√(x²+x)-√(x²-x)]·[√(x²+x)+√(x²-x)] /[√(x²+x)+√(x²-x)]
x→+∞
=lim 2x/[√(x²+x)+√(x²-x)]
x→+∞
=lim 2/[√(1+1/x)+√(1-1/x)]
x→+∞
=lim 2/(1+1)
x→+∞
=1
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