L-5

\(\displaystyle 5).\;\; \lim _{n \rightarrow \infty} \frac{1}{\sqrt[n]{n !}} \)

\(\displaystyle \Rightarrow \lim _{n \rightarrow \infty} \frac{1}{(n !)^{\frac{1}{n}}} \)

Using Cauchy's second theorem on limits which says

\(\displaystyle \{a_n\}\) of positive values,

if \(\displaystyle \lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n} = \ell\)   then  \(\displaystyle \lim_{n\rightarrow \infty} (a_n)^{1/n} = \ell\)

\(\displaystyle \text{Let }\;a_{n}=\frac{1}{n !} \)

\(\displaystyle\lim _{n \rightarrow \infty} \displaystyle \frac{a_{n+1}}{a_{n}}=\lim _{n \rightarrow \infty}\left(a_{n}\right)^{\frac{1}{n}} \)

\(\displaystyle\Rightarrow \lim _{n \rightarrow \infty} \frac{\displaystyle \frac{1}{(n+1) !}}{\frac{1}{n !}} \)

\(\displaystyle\Rightarrow \lim _{n \rightarrow \infty} \frac{1}{(n+1) !} \times \frac{n !}{1} \)

\(\displaystyle\Rightarrow \lim _{n \rightarrow \infty} \frac{1}{(n+1)}=0\)

So by Cauchy's second theorem on limits we have

\(\displaystyle \lim_{n\rightarrow \infty} (a_n)^{\frac{1}{n}} = 0\)

\(\displaystyle\text{ANSWER:}\;\; 0\)