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 $L=0$.)
 A: Notice that $L=0$ and $\log 0$ is undefined.  However, we can write
$$\lim_{n\to \infty}\left(\frac{n^2+1}{3n^2+1}\right)^n=\lim_{n\to \infty}e^{\log \left(\frac{n^2+1}{3n^2+1}\right)^n}=\lim_{n\to \infty}e^{n\log \left(\frac{n^2+1}{3n^2+1}\right)}=e^{\lim_{n\to \infty}n\log \left(\frac{n^2+1}{3n^2+1}\right)}$$
Noting that the limit in the exponent is $-\infty$, we find
$$\lim_{n\to \infty}\left(\frac{n^2+1}{3n^2+1}\right)^n=0$$
A: It's maybe easiest to note that
$$0\le{n^2+5\over 3n^2+1}\le{1\over2}\quad\text{if}\quad n\ge3$$
and therefore, by the Squeeze Theorem,
$$0\le\lim_{n\to\infty}\left(n^2+5\over 3n^2+1\right)^n\le\lim_{n\to\infty}\left(1\over2\right)^n=0$$
A: 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
$$0 < \frac{n^2+5}{3n^2+1} < 1 \implies \lim_{n \to \infty}\left( \frac{n^2+5}{3n^2+1} \right)^{\! n} = 0$$
A: Did you try solving the limit directly? I calculated the limit as $0$ as well, but through examining the dominant terms of the numerator and denominator ($n^2$ and $3n^2$), the limit of the inside of the parentheses goes to $\frac{1}{3}$, then raised to the $n^{th}$ power, goes to $0$ as $n$ goes to $\infty$
A: First, yes, your approach is correct.
Second, you could have done it more easily: if $\lim a_n = a$ and $\lim b_n = b$ (with $a,b$ not necessarily finite) and the expression $a^b$ makes sense (in the extended arithmetic that has rules for dealing with infinity), then $\lim \ a_n ^{b_n}$ exists (possibly infinite) and is equal to $a^b$. In your case, this becomes $(\frac 1 3) ^\infty = 0$.
