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