a. Đặt \(\displaystyle {u_n} = 2 + {{{{\left( { - 1} \right)}^n}} \over {n + 2}}\)
Ta có:
\(\displaystyle \eqalign{
& \left| {{u_n} - 2} \right| = {1 \over {n + 2}} < {1 \over n}\,\text{ và }\,\lim {1 \over n} = 0 \cr
& \Rightarrow \lim \left( {{u_n} - 2} \right) = 0 \Rightarrow \lim {u_n} = 2 \cr} \)
b. Đặt \(\displaystyle {u_n} = {{\sin 3n} \over {4n}} - 1\)
Ta có:
\(\displaystyle \eqalign{
& \left| {{u_n} + 1} \right| = \left| {{{\sin 3n} \over {4n}}} \right| \le {1 \over {4n}}\,\text{ và }\,\lim {1 \over {4n}} = 0 \cr
& \Rightarrow \lim \left( {{u_n} + 1} \right) = 0 \Rightarrow \lim {u_n} = - 1 \cr} \)
c. \(\displaystyle \lim {{n - 1} \over n} = \lim \left( {1 - {1 \over n}} \right) \) \(\displaystyle = \lim 1 - \lim {1 \over n} = 1\)
d. \(\displaystyle \lim {{n + 2} \over {n + 1}} = \lim {{n\left( {1 + {2 \over n}} \right)} \over {n\left( {1 + {1 \over n}} \right)}} \) \(\displaystyle = \lim {{1 + {2 \over n}} \over {1 + {1 \over n}}} = {{\lim 1 + \lim {2 \over n}} \over {\lim 1 + \lim {1 \over n}}} \) \(\displaystyle = {{1 + 0} \over {1 + 0}} = 1\)
