Giải bài 5.29 trang 87

Đề bài

Cho un=√n(√n+2−√n+1)$u_n=\sqrt{n}\left(\sqrt{n+2}-\sqrt{n+1}\right)$. Khi đó limn→+∞un$\underset{n\to +\infty }{lim}{u}_n$ bằng

A.+∞$+\infty$                         

B. 0                     

C. 12$\frac{1}{2}$                    

D. 1

Lời giải chi tiết

limn→+∞un=limn→+∞√n(√n+2−√n+1)$\underset{n\to +\infty }{lim}{u}_n=\underset{n\to +\infty }{lim}\sqrt{n}\left(\sqrt{n+2}-\sqrt{n+1}\right)$

=limn→+∞√n(√n+2−√n+1)(√n+2+√n+1)(√n+2+√n+1)$=\underset{n\to +\infty }{lim}\frac{\sqrt{n}\left(\sqrt{n+2}-\sqrt{n+1}\right)\left(\sqrt{n+2}+\sqrt{n+1}\right)}{\left(\sqrt{n+2}+\sqrt{n+1}\right)}$

=limn→+∞√n(n+2−n−1)√n+2+√n−1=limn→+∞√n√n+2+√n+1$=\underset{n\to +\infty }{lim}\frac{\sqrt{n}\left(n+2-n-1\right)}{\sqrt{n+2}+\sqrt{n-1}}=\underset{n\to +\infty }{lim}\frac{\sqrt{n}}{\sqrt{n+2}+\sqrt{n+1}}$

=limn→+∞1√1+2n+√1+2n=12$=\underset{n\to +\infty }{lim}\frac{1}{\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{2}{n}}}=\frac{1}{2}$.

Đáp án C