若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值
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若lim(2n+(an^2-2n+1)/(bn+2))=1求a/b的值若lim(2n+(an^2-2n+1)/(bn+2))=1求a/b的值若lim(2n+(an^2-2n+1)/(bn+2))=1求
若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值
若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值
若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值
纠正一下:你必须写x趋向无穷大!
可化为:
lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
lim[(2b+a)n^2+2n+1]/(bn+2)=1
因为lim(2n+(an^2-2n+1)/(bn+2))=1为一常数,
所以可知.分子分母最高阶次相等
则因为分母最高阶次为1.
所以a+2b=0
则可化为
lim(2n+1)/(bn+2))=1
上下除以2
则lim(2+1/n)/(b+2/n)=1
则b=2
所以a=-4
所以a/b=-2
a=-4,b=2,a/b=-2
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