Mình làm thử cách này nhé:
\[{u_n} = \frac{{{2^n}}}{{n!}} = \frac{{{{(1 + 1)}^n}}}{{n!}} = \frac{{C_n^0 + C_n^1 + ... + C_n^n}}{{n!}} = \frac{1}{{n!}} + \frac{1}{{(n - 1)!}} + \frac{1}{{2!(n - 2)!}} + ... + \frac{1}{{n!}}\]
Khi $n \to  + \infty $ thì $n! \to  + \infty $ suy ra $ \frac{1}{n!} \to  0 $
Vậy $\lim {u_n} = \lim \frac{1}{{n!}} + \lim \frac{1}{{(n - 1)!}} + \lim \frac{1}{{2!(n - 2)!}} + ... + \lim \frac{1}{{n!}} = 0$

\[{u_n} = \frac{{2\sin n + 3\cos n}}{{{n^2} + 1}}\]
Theo bất đẳng thức Bunhiacopsky:
\[\left| {2\sin n + 3\cos n} \right| \le \sqrt {{2^2} + {3^2}} \sqrt {{{\sin }^2}n + {{\cos }^2}n}  \le \sqrt {13} \]
\[ \Rightarrow  - \sqrt {13}  \le 2\sin n + 3\cos n \le \sqrt {13} \]
\[ \Rightarrow  - \frac{{\sqrt {13} }}{{{n^2} + 1}} \le {u_n} \le \frac{{\sqrt {13} }}{{{n^2} + 1}}\]
Ta có:
\[\lim \left( { - \frac{{\sqrt {13} }}{{{n^2} + 1}}} \right) = 0\]
\[\lim \left( {\frac{{\sqrt {13} }}{{{n^2} + 1}}} \right) = 0\]
Nên theo định lý kẹp \[\lim {u_n} = 0\]