<s>[INST] Limit of $f(n)^{g(n)}$ Given the limit:
$$\lim_ {n\to \infty}\left(\frac{n^2+5}{3n^2+1}\right)^{\! n}$$
Is it possible to assume that
$$\lim_ {n\to \infty}\left(\frac{n^2+5}{3n^2+1}\right)^{\! n} = L$$
and then take the natural log of both sides
$$\lim_ {n\to ∞}\left(n \cdot \ln\left(\frac{n^2+5}{3n^2+1}\right)\right) = \ln(L)$$
and keep solving? (I got that $L=0$.)
 [/INST] First, consider the function
$$\mathrm{f}(x) = \frac{x^2+5}{3x^2+1}$$
The derivative is given by using the quotient rule:
\begin{eqnarray*}
\mathrm{f}'(x) &=& \frac{2x(3x^2+1)-(x^2+5)(6x)}{(3x^2+1)^2} \\ \\
&=& \frac{-28}{(3x+1)^2}
\end{eqnarray*}
This tells us that $\mathrm{f}'(x) < 0$ for all $x \ge 0$, i.e. $\mathrm{f}$ is a decreasing function. This means that $$\mathrm{f}(x) < \mathrm{f}(y) \iff x > y$$
This means that the sequence $\mathrm{f}(n)$, where $n$ is a positive whole number, is strictly decreasing: 
$$\mathrm{f}(1) > \mathrm{f}(2) > \mathrm{f}(3) > \ldots > \mathrm{f}(n) > \mathrm{f}(n+1) > \ldots$$
Finally, note that when $n =2$ we have $\mathrm{f}(n) = \frac{9}{13}$, and hence $\mathrm{f}(n) < 1$ for all $n \ge 2$.
Coming back to your question. For all $n \ge 2$, we have
$$0 < \frac{n^2+5}{3n^2+1} < 1 \implies \lim_{n \to \infty}\left( \frac{n^2+5}{3n^2+1} \right)^{\! n} = 0$$
</s>