Find the limit of fraction powered by n $$\lim_{n\to\infty} \left({\frac{2n+1}{3n+2}}\right)^{n} = ?  $$ 
Simple method gives an indeterminate expression. Any idea how to think out this case?  
 A: HINT: $$0<\frac{2n+1}{3n+2}<\frac23$$ for $n\ge 1$.
A: Brian's answer is right on the money. Another way to see it is noting the algebraic equality
$$\left(\frac{2n+1}{3n+2}\right)^n=e^{n\log\frac{2n+1}{3n+2}}$$
But
$$n\log\frac{2n+1}{3n+2}\xrightarrow [n\to\infty]{}-\infty$$
since $\,\log\frac{2}{3}<0\,$ , so the limit's like 
$$\lim_{m\to-\infty}e^m=0$$
But I'd choose Brian's answer as it is way simpler...:)
A: if f(n) is less then a constant k which is less then 1, for large enough n, and all proceeding n, then as n tends to infinity $f(n)^n$=0, so the limit of your expression is zero.
A: Probably a bit cumbersome but straightforward:
$$
\lim_{n \to \infty} \bigg(\frac{2n +1}{3n+2} \bigg)^n=\lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(\frac{n+\frac{1}{2}}{n+\frac{2}{3}} \bigg)\bigg)^n = \lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(\frac{n+\frac{2}{3}-\frac{1}{6}}{n+\frac{2}{3}}\bigg)\bigg)^n
=\lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(1-\frac{1}{6n+4} \bigg)\bigg)^n= \lim_{n \to \infty}\bigg( \frac{2}{3} \bigg)^n \lim_{n \to \infty} \bigg(1-\frac{1}{6n+4} \bigg)^{(6n +4-4)\cdot \frac{1}{6}}=0 \cdot e^{-\frac{1}{6}} \cdot 1 
= 0
$$
