Giải bài 7 trang 122 – SGK môn Đại số và Giải tích lớp 11

a)

 $\begin{aligned} & \lim \left( {{n}^{3}}+2{{n}^{2}}-n+1 \right)=\lim {{n}^{3}}\left( 1+\dfrac{2}{n}-\dfrac{1}{{{n}^{2}}}+\dfrac{1}{{{n}^{3}}} \right) \\ & =\lim {{n}^{3}}.\lim \left( 1+\dfrac{2}{n}-\dfrac{1}{{{n}^{2}}}+\dfrac{1}{{{n}^{3}}} \right)=+\infty \\ \end{aligned}$

Vì $\lim {{n}^{3}}=+\infty$  và $\lim \left( 1+\dfrac{2}{n}-\dfrac{1}{{{n}^{2}}}+\dfrac{1}{{{n}^{3}}} \right)=1>0$

b)

$\begin{aligned} & \lim \left( -{{n}^{2}}+5n-2 \right)=\lim {{n}^{2}}\left( -1+\dfrac{5}{n}-\dfrac{2}{{{n}^{2}}} \right) \\ & =\lim {{n}^{2}}.\lim \left( -1+\dfrac{5}{n}-\dfrac{2}{{{n}^{2}}} \right)=-\infty \\ \end{aligned}$

Vì $\lim {{n}^{2}}=+\infty$ và $\lim \left( -1+\dfrac{5}{n}-\dfrac{2}{{{n}^{2}}} \right)=-1<0$

c)

$\begin{aligned} & \lim \left( \sqrt{{{n}^{2}}-n}-n \right)=\lim \dfrac{\left( {{n}^{2}}-n \right)-{{n}^{2}}}{\sqrt{{{n}^{2}}-n}+n}=\lim \dfrac{-n}{n\left( \sqrt{1-\dfrac{1}{n}}+1 \right)} \\ & =\lim \dfrac{-1}{\sqrt{1-\dfrac{1}{n}}+1}=\dfrac{-1}{2} \\ \end{aligned}$

d)

$\lim \left( \sqrt{{{n}^{2}}-n}+n \right)=\lim n\left( \sqrt{1-\dfrac{1}{n}}+1 \right)=+\infty$

Vì $\lim n=+\infty$  và $\lim \left( \sqrt{1-\dfrac{1}{n}}+1 \right)=2>0$