evaluating the limit $\lim \limits_{n \to \infty} \frac{(4(n*3^n + 3))^n}{(3^{n+1} (n+1)+3)^{n+1}}$ I'm trying to solve the following limit: $\lim \limits_{n \to \infty} \frac{(4(n3^n + 3))^n}{(3^{n+1} (n+1)+3)^{n+1}}.$ I've got no idea where to even start, it's just too big! I don't know whether L'hospital would be a good idea considering, well, look at those functions. I also don't really see a nice way to apply the squeeze theorem. 
Maybe it would be a good idea to prove that for all $n \in \mathbb{N}$, $4(n3^n + 3)$ is smaller than $3^{n+1} (n+1)+3$ so that we can conclude that the limit is zero or something, but even that induction proof would be a monster (I tried it).
 A: Begin by noticing that for the denominator the $3^{n+1}(n+1)$ term dominates the $+3$ term in the limit. 
Show that $\lim_{n \to \infty} (3^{n+1}(n+1)+3)^{n+1} = \lim_{n \to \infty} (3^{n+1}(n+1))^{n+1}$
You can use this idea to remove the addition in the numerator and denominator
A: We have that:
$$\lim \limits_{n \to \infty} \frac{(4(n3^n + 3))^n}{(3^{n+1} (n+1)+3)^{n+1}}$$
First, look at the denominator:
$$(3^{n+1} (n+1)+3)^{n+1}$$
Note that the term $3^{n+1} (n+1)$ will be much larger than the term $3$ as $n\to\infty$. Therefore, we can simplify and say that the denominator turns into:
$$(3^{n+1} (n+1)+3)^{n+1} \to (3^{n+1} (n+1))^{n+1}$$
Similarly, we can show in the numerator that:
$$(4(n3^n + 3))^n \to (4n3^n)^n$$
Our limit then simplifies to:
$$\lim \limits_{n \to \infty} \frac{(4n3^n)^n}{(3^{n+1} (n+1))^{n+1}} = \lim \limits_{n \to \infty} \frac{4^{n}n^{n}3^{n^2}}{3^{(n+1)^{2}}\left(n+1\right)^{n+1}}$$
Note the terms in the denominator is $n+1$, and the terms in the numerator is $n$.
Can you take it from here?
A: With equivalents:


*

*$(n3^n+3)^n\sim_\infty (n3^n)^n=n^n 3^{n^2}$ (the difference of the logs tends to $0$).

*$(3^{n+1} (n+1)+3)^{n+1}\sim_\infty (3^{n+1} (n+1))^{n+1}=3^{(n+1)^2} (n+1)^{n+1} $


Hence
\begin{align*}\frac{(4(n*3^n + 3))^n}{(3^{n+1} (n+1)+3)^{n+1}}\sim_\infty&\frac{4^n3^{n^2}n^n}{3^{(n+1)^2} (n+1)^{n+1}}=\frac{4^n }{3^{2n+1}} \biggl(\frac{n}{n+1}\biggr)^n\frac1{n+1}\\ 
&=\biggl(\frac49\biggr)^n\frac1{3(n+1)}\biggl(\frac{n}{n+1}\biggr)^n\to 0\cdot 0\cdot\frac1{\mathrm e}=0.
\end{align*}
