More and more limits for sequences So here goes a bit of homework:
$$\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}$$
Well, this would trivially lead to:
$$\lim_{n\to\infty}{\left(\frac{3+\frac{2}{n}+\frac{1}{n^2}}{3+\frac{5}{n^2}}\right)^{\frac{n\left(1+\frac{2}{n^2}\right)}{2+\frac{1}{n}}}}$$
Which is clearly an indetermination of type "$1^\infty$". Now, I can't really get through this step... Any hints?
 A: The standard trick for dealing with $1^\infty$ forms is to take logs; it’s very useful if you don’t see anything slicker. Let 
$$L=\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\;;$$
then 
$$\begin{align*}
\ln L&=\ln\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\\
&=\lim_{n\to\infty}\ln{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\\
&=\lim_{n\to\infty}\left(\frac{n^2+2}{2n+1}\right)\ln\left(\frac{3n^2+2n+1}{3n^2-5}\right)\;,
\end{align*}$$
where the second step uses the continuity of the log function. This is an $\infty\cdot 0$ form, which you can easily convert to a $\frac00$ form:
$$\lim_{n\to\infty}\frac{\ln\left(\frac{3n^2+2n+1}{3n^2-5}\right)}{\frac{2n+1}{n^2+2}}\;.$$
Once you know $\ln L$, recovering $L$ is trivial; just remember to do it!
A: Reduce to the limit for the exponential:
$$
   \lim_{n \to \infty} \left( 1 + \frac{2}{3 n}\frac{1 - \frac{2}{n}}{1 + \frac{5}{3n}} \right)^{\frac{n}{2} - \frac{n-4}{4n+2}} = \underbrace{\lim_{n \to \infty} \left( 1+\frac{2}{3n} \right)^{n/2}}_{\exp\left(\frac{1}{3}\right)} \cdot \underbrace{\lim_{n \to \infty} \left(1 + \frac{2}{3 n}\frac{1 - \frac{2}{n}}{1 + \frac{5}{3n}}\right)^{-\frac{n-4}{4n+2}} }_{1} \cdot  \lim_{n \to \infty} \left(\frac{1 + \frac{2}{3 n}\frac{1 + \frac{2}{n}}{1 + \frac{5}{3n}}}{1-\frac{2}{3n}}\right)^{n/2}
$$
Using 
$$
  \frac{1 + \frac{2}{3 n}\frac{1 + \frac{2}{n}}{1 + \frac{5}{3n}}}{1-\frac{2}{3n}} = 1 - \frac{22}{(3n+2)(3n+5)}
$$
we conclude that the last limit also equals to 1.
A: $$\frac{3n^2+2n+1}{3n^2-5}=1+\frac{2n+6}{3n^2-5}=1+\frac{2}{\frac{3n^2-5}{n+3}}\Longrightarrow$$
$$\Longrightarrow\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}=\left[\left(1+\frac{2}{\frac{3n^2-5}{n+3}}\right)^{\frac{3n^2-5}{n+3}}\right]^\frac{n^3+3n^2+2n+6}{6n^3+3n^2-10n-5}\Longrightarrow$$
Well, now the inner limit must be well-known, and for the exterior one just check the exponent behaves as $\,1/6\,$ for large values of $\,n\,$ . The limit indeed is $\,e^{1/3}\,$
