Toán lim 1/n-1 =……….. lim (4-3n ²)/n+1 =…….. lim (n ³/3 -2n +2/3)=……….. 06/08/2021 By Caroline lim 1/n-1 =……….. lim (4-3n ²)/n+1 =…….. lim (n ³/3 -2n +2/3)=………..
$\lim\dfrac{1}{n-1}=\lim\dfrac{\dfrac{1}{n}}{1-\dfrac{1}{n}}=0$ $\lim\dfrac{4-3n^2}{n+1}=\lim\dfrac{n^2\Big(\dfrac{4}{n^2}-3\Big)}{n\Big(1+\dfrac{1}{n}\Big)}=\lim n.\dfrac{\dfrac{4}{n^2}-3}{1+\dfrac{1}{n}}=-\infty$ $\lim\Big(\dfrac{n^3}{3}-2n+\dfrac{2}{3}\Big)=\lim n^3\Big(\dfrac{1}{3}-\dfrac{2}{n^2}+\dfrac{2}{3n^3}\Big)=+\infty$ Trả lời
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$\lim\dfrac{1}{n-1}=\lim\dfrac{\dfrac{1}{n}}{1-\dfrac{1}{n}}=0$
$\lim\dfrac{4-3n^2}{n+1}=\lim\dfrac{n^2\Big(\dfrac{4}{n^2}-3\Big)}{n\Big(1+\dfrac{1}{n}\Big)}=\lim n.\dfrac{\dfrac{4}{n^2}-3}{1+\dfrac{1}{n}}=-\infty$
$\lim\Big(\dfrac{n^3}{3}-2n+\dfrac{2}{3}\Big)=\lim n^3\Big(\dfrac{1}{3}-\dfrac{2}{n^2}+\dfrac{2}{3n^3}\Big)=+\infty$