$\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?
$\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?
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 $
HINT: $0<\frac{2n+1}{3n+2}<\frac23$ for $n\ge 1$.
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...:)
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.