Đáp án:
*lim$\sqrt[3]{1+2n-n^{3}}$=-∞
*lim$\frac{n^{2}+1}{2n^{4}+n+1}$=0
*Un=$\frac{3^{n}+1}{2^{n}-1}$ là dãy tăng.
Giải thích các bước giải:
*lim$\sqrt[3]{1+2n-n^{3}}$=lim($n^{3}$.$\sqrt[3]{\frac{1}{n^{3}}+\frac{2}{n^{2}}-1}$)=lim(-$n^{3}$)=-∞
*lim$\frac{n^{2}+1}{2n^{4}+n+1}$=lim$\frac{n^{2}.(1+\frac{1}{n^{2}})}{n^{4}.(2+\frac{1}{n^{3}}+\frac{1}{n^{4}})}$=lim$\frac{1}{n^{2}}$=0
*Un=$\frac{3^{n}+1}{2^{n}-1}$
Ta có:
Un+1-Un=$\frac{3^{n+1}+1}{2^{n+1}-1}$-$\frac{3^{n}+1}{2^{n}-1}$>0
