$\lim_{} \frac{-3n^{2}+4n+1}{n^{2}. 2^{n} }$ 30/10/2021 Bởi Josephine $\lim_{} \frac{-3n^{2}+4n+1}{n^{2}. 2^{n} }$
$I=\lim\dfrac{-3n^2+4n+1}{n^2.2^n}$ $=\lim\dfrac{-3+\dfrac{4}{n}+\dfrac{1}{n^2}}{2^n}$ $\lim\Big(-3+\dfrac{4}{n}+\dfrac{1}{n^2}\Big)=-3$ $\lim 2^n=+\infty$ $\to I=0$ Bình luận
`lim (-3n^2 + 4n + 1)/(n^{2}.2^{n})` $= lim \dfrac{-3 + \dfrac{4}{n} + \dfrac{1}{n^2}}{1.2^{n}}$ $= \dfrac{lim (-3 + \dfrac{4}{n} + \dfrac{1}{n^2})}{lim 2^{n}}$ `= -3/(+\infty)` `= 0` Bình luận
$I=\lim\dfrac{-3n^2+4n+1}{n^2.2^n}$
$=\lim\dfrac{-3+\dfrac{4}{n}+\dfrac{1}{n^2}}{2^n}$
$\lim\Big(-3+\dfrac{4}{n}+\dfrac{1}{n^2}\Big)=-3$
$\lim 2^n=+\infty$
$\to I=0$
`lim (-3n^2 + 4n + 1)/(n^{2}.2^{n})`
$= lim \dfrac{-3 + \dfrac{4}{n} + \dfrac{1}{n^2}}{1.2^{n}}$
$= \dfrac{lim (-3 + \dfrac{4}{n} + \dfrac{1}{n^2})}{lim 2^{n}}$
`= -3/(+\infty)`
`= 0`