$\lim\limits_{n \to \infty} \frac{n 2^{n}}{3^{n}}$. without L'Hôpital's rule I have the following limit for which I cannot use L'Hôpital's rule:
$$\lim\limits_{n \to \infty} \frac{n 2^{n}}{3^{n}}$$
The only thing that I know is that $\lim\limits_{x \to \infty}\, \frac{x^{\alpha}}{a^{x}} = 0$ when $x > 0$ and $a > 1$.
I'm aware that the limit is 0, but I can't figure out how to calculate that, I tried to squeeze theorem but I'm stuck with:
$$0 < \frac{n 2^{n}}{3^{n}} < $$
and I can't find another term that goes to zero that is bigger than the term in the middle.
I understand the question can be trivial but I've no problem with other limits. Thanks in advance for your insights.
 A: Here is another way to approach it for the sake of curiosity.
According to the ratio test, one has that
\begin{align*}
\limsup_{n\to\infty}\frac{a_{n+1}}{a_{n}} = \limsup_{n\to\infty}\frac{2}{3}\left(\frac{n+1}{n}\right) = \frac{2}{3} < 1
\end{align*}
Consequently, the following series converges:
\begin{align*}
\sum_{n=1}^{\infty}\frac{n2^{n}}{3^{n}} = 1\times\frac{2}{3} + 2\times\frac{4}{9} + 3\times\frac{8}{27} + \ldots
\end{align*}
By its turn, this means that its general term $a_{n}$ converges to zero, and we are done.
Hopefully this helps!
A: 
The only thing that I know is that $\lim_{x \to \infty}\, \dfrac{x^{\alpha}}{a^{x}} = 0$ when $x > 0$ and $a > 1$.

Then use $$\frac{n2^n}{3^n} = \frac{n^1}{(3/2)^n}$$
A: Using binomial theorem:
$$\left(\frac{3}{2}\right)^n=
\left(1+\frac{1}{2}\right)^n=
\sum\limits_{k=0}^n\binom{n}{k}\left(\frac{1}{2}\right)^k>
1+\frac{n}{2}+\frac{n(n-1)}{2}\cdot\frac{1}{4}$$
then
$$0<\frac{n2^n}{3^n}=\frac{n}{\left(\frac{3}{2}\right)^n}<\frac{n}{1+\frac{n}{2}+\frac{n(n-1)}{8}}=\frac{1}{\frac{1}{n}+\frac{1}{2}+\frac{n-1}{8}}\to0, n\to\infty$$
A: Bernoulli's inequality states that $(1+x)^n \ge nx$ for $x > -1$ and $n \ge 1$. This is easily proved by induction. In particular, if $x > 0$ then $$\left( \frac 1{x+1} \right)^n \le \frac 1{nx}.$$
If you try to get a $\frac 23$ on the left you will take $x = \frac 12$ which gives you $n \left( \dfrac 23 \right)^n \le 2$.  Not exactly what you wanted, but at least it shows the sequence is bounded.
But, if you do the same thing with a smaller value of $x$, say $x = \frac 13$, you obtain $$n \left( \dfrac 34 \right)^n \le 3$$ which leads immediately to
$$n \left( \dfrac 23 \right)^n \le 3 \left( \dfrac 89 \right)^n$$ and now you can apply the squeeze theorem.
