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: $n \in \mathbb{N}$
 A: 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:
$$\frac{\frac{n^4}{4^n}+1}{\frac{n}{4^{n}}+4} \to \frac{0+1}{0+4}=\frac{1}{4}$$
A: Hint:
$$
\frac{n^4+4^n}{n+4^{n+1}}=\frac{\frac{n^4}{4^n}+1}{\frac{n}{4^n}+4}
$$
A: As you said you can do $$\lim_{n\rightarrow\infty} \frac{n^4+4^n}{n+4^{n+1}} = \lim_{n\rightarrow\infty} \frac{\frac{n^4}{4^n}+1}{\frac{n}{4^n}+4} =  \frac{\lim_{n\rightarrow\infty} \frac{n^4}{4^n}+1}{\lim_{n\rightarrow\infty} \frac{n}{4^n}+4} = \frac{0+1}{0+4}$$
But it's necessary that you already have proven in you lecture, that


*

*$\lim_{n\rightarrow\infty}\frac{n^k}{q^n} = 0$ when $|q| > 1$ and $k\in\mathbb N$

*$\lim_{n\rightarrow\infty}\frac{a_n}{b_n} = \frac{\lim_{n\rightarrow\infty} a_n}{\lim_{n\rightarrow\infty}b_n}$ when $(a_n)$ and $(b_n)$ are convergent

*$\lim_{n\rightarrow\infty}(a_n+b_n) = \lim_{n\rightarrow\infty} a_n+\lim_{n\rightarrow\infty}b_n$ when $(a_n)$ and $(b_n)$ are convergent


I guess you will find similar theorems in your script...
A: exponential function $4^n$ is greater than any power of $n$ for $n$ large. that is $\lim_{n \to \infty}\dfrac{n^k}{4^n} = 0$   for any $k.$  assuming this,
$\dfrac{n^4 + 4^n}{n + 4^{n+1}} = \dfrac{ 4^n + \cdots}{4^{n+1} + \cdots} = \dfrac{1}{4} + \cdots$
establishing the limit. suppose $\lim_{n \to \infty}\dfrac{n^k}{4^n} = L$  replacing $n$ by $2n$ we shoud get the same limit. therefore 
 $$L = \lim_{n \to \infty}\dfrac{n^k}{4^n} = \lim_{n \to \infty}\dfrac{(2n)^k}{4^{2n}} = \lim_{n \to \infty}\frac{2^k}{4^n}\dfrac{n^k}{4^n}= \lim_{n \to \infty}\frac{2^k}{4^n}L = 0.$$
