Limit of $\frac{n^4+4^n}{n+4^{n+1}}$ Just when i  thought i finally got the hang of limits, i stumbled upon this:
$$\frac{n^4+4^n}{n+4^{n+1}}$$
Now, this kinda makes sense in my head because $4^n$ grows  a lot faster than $n^4$, let alone $n$. Now my question is if this assumption is a valid tool for solving this limit i.e. if i divide both denominator and numerator with $4^n$, can that be user without further explanation or proof. 
This, of course, gives the result of $\frac{1}{4}$ which seems to be correct.
EDIT: $n \in \mathbb{N}$

Note that:
$$\frac{n^4+4^n}{n+4^{n+1}}=\frac{4^{-n}}{4^{-n}}\frac{n^4+4^n}{n+4^{n+1}}=\frac{\frac{n^4}{4^n}+1}{\frac{n}{4^{n}}+4}$$
Now you have:
$$\frac{n^4}{4^n} \to 0$$
$$\frac{n}{4^n} \to 0$$
when $n \to \infty$, so:
$$\frac{\frac{n^4}{4^n}+1}{\frac{n}{4^{n}}+4} \to \frac{0+1}{0+4}=\frac{1}{4}$$
