$\lim _{n \rightarrow \infty}\left(1+\frac{e^{n}}{5^{n}}\right)^{3 n^{2}+1}$

Step 1 ：Given the expression \(\lim _{n \rightarrow \infty}\left(1+\frac{e^{n}}{5^{n}}\right)^{3 n^{2}+1}\)

Step 2 ：This expression is in the form of \(\lim _{n \rightarrow \infty}\left(1+\frac{a}{n}\right)^{bn}\), which is a well-known limit that equals \(e^{ab}\)

Step 3 ：In this case, \(a = e^n\) and \(b = 3n^2+1\). However, we need \(a\) to be in the form of \(\frac{1}{n}\), so we need to adjust the expression a bit

Step 4 ：We can rewrite the expression as \(\lim _{n \rightarrow \infty}\left(1+\frac{1}{\frac{5^n}{e^n}}\right)^{3 n^{2}+1}\)

Step 5 ：Now, \(a = \frac{1}{\frac{5^n}{e^n}} = \frac{e^n}{5^n}\) and \(b = 3n^2+1\)

Step 6 ：So, the limit should be \(e^{ab} = e^{(\frac{e^n}{5^n})(3n^2+1)}\)

Step 7 ：However, as \(n\) approaches infinity, \(\frac{e^n}{5^n}\) approaches 0 because \(e^n\) grows slower than \(5^n\)

Step 8 ：Therefore, the limit should be \(e^{0*(3n^2+1)} = e^0 = 1\)

Step 9 ：Final Answer: \(\boxed{1}\)