1.lim 2n^3-n-3/5n+1 2. Lim (2n-1)^3(n-3)^5/4n^8+2n-4

Đáp án:

$1)$ $\lim \dfrac{2n^3-n-3}{5n+1}$  $=+\infty \:$

$2)$ $\lim \dfrac{\left(2n-1\right)^3\left(n-3\right)^5}{4n^8+2n-4}$ $=2$

Giải thích các bước giải:

$1)$ $\lim \dfrac{2n^3-n-3}{5n+1}$

$=\lim \dfrac{2n^2-1-\dfrac{3}{n}}{5+\dfrac{1}{n}}$

$=\dfrac{\lim 2n^2-1-\dfrac{3}{n}}{\lim \:5+\dfrac{1}{n}}$

$=\dfrac{\infty \:}{5}$

$=+\infty \:$

$2)$ $\lim \dfrac{\left(2n-1\right)^3\left(n-3\right)^5}{4n^8+2n-4}$

$=\lim \dfrac{\left(n\left(2-\dfrac{1}{n}\right)\right)^3\left(n\left(1-\dfrac{3}{n}\right)\right)^5}{n^8\left(4+\dfrac{2}{n^7}-\dfrac{4}{n^8}\right)}$

$=\lim \:\dfrac{\left(-\dfrac{1}{n}+2\right)^3\left(-\dfrac{3}{n}+1\right)^5}{4+\dfrac{2}{n^7}-\dfrac{4}{n^8}}$

$=\dfrac{\lim \left(-\dfrac{1}{n}+2\right)^3\left(-\dfrac{3}{n}+1\right)^5}{\lim \:4+\dfrac{2}{n^7}-\dfrac{4}{n^8}}$

$=\dfrac{8}{4}$

$=2$

Đáp án:

$\begin{array}{l} 1)\lim \dfrac{{2{n^3} - n - 3}}{{5n + 1}}\\  = \lim \dfrac{{{n^3}\left( {2 - \dfrac{1}{{{n^2}}} - \dfrac{3}{{{n^3}}}} \right)}}{{n.\left( {5 + \dfrac{1}{n}} \right)}}\\  = \lim \dfrac{{{n^2}\left( {2 - \dfrac{1}{{{n^2}}} - \dfrac{3}{{{n^3}}}} \right)}}{{5 + \dfrac{1}{n}}}\\  =  + \infty \\ 2)\lim \dfrac{{{{\left( {2n - 1} \right)}^3}.{{\left( {n - 3} \right)}^5}}}{{4{n^8} + 2n - 4}}\\  = \lim \dfrac{{\dfrac{{{{\left( {2n - 1} \right)}^3}}}{{{n^3}}}.\dfrac{{{{\left( {n - 3} \right)}^5}}}{{{n^5}}}}}{{4 + \dfrac{2}{{{n^7}}} - \dfrac{4}{{{n^8}}}}}\\  = \lim \dfrac{{{{\left( {2 - \dfrac{1}{{{n^3}}}} \right)}^3}.{{\left( {1 - \dfrac{3}{n}} \right)}^5}}}{{4 + \dfrac{2}{{{n^7}}} - \dfrac{4}{{{n^8}}}}}\\  = \dfrac{{{2^3}{{.1}^5}}}{4}\\  = \dfrac{8}{4}\\  = 2 \end{array}$