a. Ta có:
\(\left| {0,99} \right| < 1\,\text{ nên }\,\lim {u_n} = \lim {\left( {0,99} \right)^n} = 0\)
b.
\(\eqalign{
& \left| {{u_n}} \right| = \left| {{{{{\left( { - 1} \right)}^n}} \over {{2^n} + 1}}} \right| = {1 \over {{2^n} + 1}} < {\left( {{1 \over 2}} \right)^n},\lim {\left( {{1 \over 2}} \right)^n} = 0 \cr
& \Rightarrow \lim {u_n} = 0 \cr} \)
c.
\(\eqalign{
& \left| {{u_n}} \right| = {{\left| {\sin {{n\pi } \over 5}} \right|} \over {{{\left( {1,01} \right)}^n}}} \le {\left( {{1 \over {1,01}}} \right)^n},\lim {\left( {{1 \over {1,01}}} \right)^n} = 0 \cr
& \Rightarrow \lim {u_n} = 0 \cr} \)