Giải Bài tập 5.1 trang 109 sgk Toán 11 tập 1 Kết nối

a) $\underset{n\rightarrow +\infty }{lim}\frac{n^{2}+n+1}{2n^{2}+1}=\underset{n\rightarrow +\infty }{lim}\frac{1+\frac{1}{n}+\frac{1}{n^{2}}}{2+\frac{1}{n^{2}}}=\frac{\underset{n\rightarrow +\infty }{lim}(1+\frac{1}{n}+\frac{1}{n^{2}})}{\underset{n\rightarrow +\infty }{lim}(2+\frac{1}{n^{2}})}=\frac{1}{2}$

b) $\underset{n\rightarrow +\infty }{lim}v_{n}=\underset{n\rightarrow +\infty }{lim}(\sqrt{2n^{2}+1}-n)=\underset{n\rightarrow +\infty }{lim}\frac{n^{2}+2n-n^{2}}{\sqrt{n^{2}+2n}+n}=\underset{n\rightarrow +\infty }{lim}\frac{2n}{n(\sqrt{1+\frac{2}{n}}+1)}=\underset{n\rightarrow +\infty }{lim}\frac{2}{\sqrt{1+\frac{2}{n}}+1}=1$