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lim(n趋向无穷)=[(5x^2+1)/(3x-1)]sin1/x=?
人气:425 ℃ 时间:2020-04-02 03:05:32
解答
lim(n趋向无穷)[(5x^2+1)/(3x-1)]sin1/x
=lim(n趋向无穷)[(5x^2+1)/(3x-1)*1/x
=lim(n趋向无穷)(5x^2+1)/(3x^2-x)
=lim(n趋向无穷)(5+1/x^2)/(3-1/x)
=(5+0)/(3-0)
=5/3
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