Which function grows faster? Which function grows faster 
$()= 2^{^2+3}$ and $() = 2^{+1}$
by using the limit theorem I will first simplify 
then I will just get $$\lim_{n \to \infty} \dfrac{2^{n^2+3n}}{2^{n+1}}=\lim_{n \to \infty} 2^{n^2+3n-n-1}=\lim_{n \to \infty} 2^{n^2+2n-1}=\infty$$
Is this enough? 
I say it will go then to infinity so the $f(n)$ is growing faster? I am asking this question because I have to find it by using limit but I didn't need to use l'hopital rule! 
 A: Before Edit:
Your idea was correct, but you didn’t simplify the limit properly.
$$\lim_{n \to \infty} \frac{2^{n^2+3n}}{2^n+1}$$
It is enough to divide both the numerator and denominator by $2^n$.
$$\lim_{n \to \infty} \frac{\frac{2^{n^2+3n}}{2^n}}{\frac{2^n+1}{2^n}} = \lim_{n \to \infty} \frac{2^{n^2+3n-n}}{2^{n-n}+\frac{1}{2^n}} = \lim_{n \to \infty} \frac{2^{n^2+2n}}{1+\frac{1}{2^n}}$$
As $n \to \infty$, it becomes clear that the limit tends to $\infty$ since the numerator tends to $\infty$ while the denominator tends to $1$.
After Edit: Yes, your way is correct.
A: HINT
You conclusion is correct but that step is wrong
$$\lim_{n \to \infty}= \dfrac{2^{n^2+3n}}{2^n+1}\color{red}{=\lim_{n \to \infty} 2^{n^2+3n-n-1}}$$
you could use that $2^n+1\le 2^{n+1}$ and therefore
$$\dfrac{2^{n^2+3n}}{2^n+1}\ge \dfrac{2^{n^2+3n}}{2^{n+1}}$$
Update after editing
For $g(n)=2^{n+1}$ your method is absolutely correct.
What about $f(n)=2^{n^2+3n}$ and $g(n)=3^{n+1}$?
A: It is $$\frac{2^{n^2}\cdot 2^{3n}}{2^n\left(1+\frac{1}{2^n}\right)}=\frac{2^{n^2+2n}}{1+\frac{1}{2^n}}$$
