limx→∞[(x+1)/(x+2)]^x
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limx→∞[(x+1)/(x+2)]^xlimx→∞[(x+1)/(x+2)]^xlimx→∞[(x+1)/(x+2)]^x(x+1)/(x+2)=1-1/(x+2)所以令1/a=-1/(x+2)x
limx→∞[(x+1)/(x+2)]^x
limx→∞[(x+1)/(x+2)]^x
limx→∞[(x+1)/(x+2)]^x
(x+1)/(x+2)=1-1/(x+2)
所以令1/a=-1/(x+2)
x=-a-2
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