Evaluate the limit $\lim_{n\to\infty} \left(\frac{3n}{3n + 1}\right)^n$ I've been trying to evaluate this limit and can't seem to find a way around.
I know how to show that
$\lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n = e$ but I can't find a way to apply it.
 A: Look at the limit of the reciprocal; it looks a lot more like the limit you know.
A: Let's apply just what you know
$$\lim_{n\to+\infty}\left(\frac{3n}{3n+1}\right)^n=\lim_{n\to+\infty}\left(1+\frac{1}{3n}\right)^{-n}=\lim_{n\to+\infty}\left(\left(1+\frac{1}{3n}\right)^{3n}\right)^{-1/3}\\=\left(\lim_{\alpha\to+\infty}\left(1+\frac{1}{\alpha}\right)^{\alpha}\right)^{-1/3}=e^{-1/3}$$
A: Let
$$
y_n=\left(\frac{3n}{3n+1}\right)^n
$$
so that
$$
\ln(y_n)=n\log\left(\frac{3n}{3n+1}\right).
$$
Can you use L'Hopital's rule to find the limit of $\ln(y_n)$ as $n\rightarrow\infty$? If so, you can use $y_n=e^{\ln(y_n)}$, along with continuity of the exponential function, to translate this in to a limit for $y_n$.
A: Hint:
$$
 \left(\frac{3n}{3n + 1}\right)^n = \left(1-\frac{1}{3n + 1}\right)^n= \left(\left(1-\frac{1}{3n + 1}\right)^{3n+1}\cdot\frac{3n + 1}{3n}\right)^{1/3}
$$
Since the function $f\colon x\geq 0\mapsto \sqrt[3]x$ is continuous, if you show that the quantity inside the outer parentheses converges to $\ell\geq 0$, the overall quantity will converge to $\sqrt[3]\ell$.
A: A simple comparison yields:
$$\left(1-\dfrac{1}{3n}\right)^n\le\left(\dfrac{3n}{3n+1}\right)^n\le\left(1-\dfrac{1}{3n+3}\right)^{n+1}\dfrac{3n+3}{3n+2}.$$
Now both the left and the right side converge to $e^{-1/3}$.  
A: 
there's a general result you can use if $\lim_{x\to \infty}f(x)=0$ and $\lim_{x \to \infty}g(x)=\infty$ then $\lim_{x\to \infty}\left( 1+f(x) \right)^{g(x)}=e^{\lim_{x\to \infty}f(x)g(x)}$
  Here write  $\lim_{n\to\infty} \left(\frac{3n}{3n + 1}\right)^n=\lim_{x\to\infty}\left(1-\frac{1}{3n+1}\right)^{n}=e^{\lim_{x\to \infty}\frac{-n}{3n+1}}=e^{\frac{-1}{3}}$

