comptuting $ \lim\limits_{n\to\infty} \frac{3^{n}+\sqrt{n}}{n!+2^{(n+1)}}$ I have been trying to solve this limit for more than an hour and I'm stuck.

$$ \lim_{n\to\infty} \frac{3^{n}+\sqrt{n}}{n!+2^{(n+1)}}$$

What I've so far is:
$$ \lim_{n\to\infty} \frac{3^{n}+\sqrt{n}}{n!+2^{(n+1)}}\\ \lim_{n\to\infty} \frac{3^{n}}{n!+2^{(n+1)}}+\lim_{n\to\infty} \frac{\sqrt{n}}{n!+2^{(n+1)}}\\\lim_{n\to\infty} \frac{3^{n}}{1+\frac{2^{(n+1)}}{n!}}\frac{1}{n!}+\lim_{n\to\infty} \frac{\sqrt{n}}{1+\frac{2^{(n+1)}}{n!}}\frac{1}{n!}\\ \lim_{n\to\infty} \frac{3^{n}}{1+\frac{2^{(n+1)}}{n!}}*0+\lim_{n\to\infty} \frac{\sqrt{n}}{1+\frac{2^{(n+1)}}{n!}}*0\\ \lim_{n\to\infty} \frac{3^{n}}{1+0}*0+\lim_{n\to\infty} \frac{\sqrt{n}}{1+0}*0\\ \lim_{n\to\infty} \frac{3^{n}}{1}*0+\lim_{n\to\infty} \frac{\sqrt{n}}{1}*0\\ \lim_{n\to\infty}{3^{n}}*0+\lim_{n\to\infty} \sqrt{n}*0\\ \lim_{n\to\infty}{\inf}*0+\lim_{n\to\infty} \inf*0$$
Which is an undetermination. 
And I don't know how to continue it.
Can someone help me?
Please let me know if something isn't very clear. Thank you
 A: You can divide the limit in two pieces but can't take the limit only for a single part as you have done here
$$\color{red}{\lim_{n\to\infty} \frac{3^{n}}{1+\frac{2^{(n+1)}}{n!}}*0+\lim_{n\to\infty} \frac{\sqrt{n}}{1+\frac{2^{(n+1)}}{n!}}*0}$$
The limit can be easily handled as follow
$$\frac{3^{n}+\sqrt{n}}{n!+2^{(n+1)}}=\frac{3^{n}}{n!}\frac{ 1 +\frac{\sqrt{n}}{3^{n}}}{1+\frac{2^{(n+1)}}{n!}}\to0\cdot\frac{1+0}{1+0}=0$$
indeed by ratio test
$$\frac{3^{n}}{n!}\to 0 \quad \quad \frac{2^{(n+1)}}{n!} \to0 \quad \quad \frac{\sqrt{n}}{3^{n}}\to0$$
A: Hint: Compare your original fraction to $\frac{2\cdot 3^n}{n!}$, which is a lot easier to work with.
A: My advice: don't systematically replace part of an expression by its limit as soon as you know it, while keeping the $n$ in the rest of the expression. 
For instance when you have $\sqrt{n} \dfrac{1}{n!} $, you rush into replacing $\dfrac{1}{n!} $ by $0$ and then you're stuck. Analyse the situation: $\sqrt{n}$ is actually a very slow sequence compared to $n!$. You can write $\sqrt{n} \dfrac{1}{n!} \leq n\dfrac{1}{n!} =\dfrac{1}{(n-1)!} $ which solves the problem!
Also, be very careful when you have a sum, like in your first step: it may happen that $\lim (u_n+v_n)$ exists but  neither of $\lim (u_n)$ and $\lim (v_n)$ does. In other words $u_n$ needs $+v_n$ to converge. In that case your cause would be lost from the very first step. However here you are dealing with positive sequences only, which prevents this phenomenon to happen.
A: 
we have 
$$ \frac{3^{n}+\sqrt{n}}{n!+2^{(n+1)}}\le  \frac{3^{n}+\sqrt{n}}{n!} =\frac{3^{n}}{n!} + \frac{1}{\sqrt{n}(n-1)!}\to0$$
Indeed, the convergence of the series $$e^3 =\sum_{n=0}^{\infty}\frac{3^{n}}{n!}\implies \frac{3^{n}}{n!} \to0$$
