Showing that $\displaystyle\lim_{n\rightarrow \infty} (4^{n}-n^{4})= \infty.$ I can think of a number of ways to prove that $\displaystyle\lim_{n\rightarrow \infty} (4^{n}-n^{4})= \infty$ but none of them brief. Does anyone have a short proof of this?
Thanks!
 A: Denote $a_n = 4^n$ and $b_n = n^4$. Clearly, $a_4 = b_4$ and $a_6>2b_6$. Suppose that for some $n\geq 6$ it holds $a_n\geq 2b_n$ then
$$
a_{n+1}=4a_n\geq 8b_n=8\left(\frac{n}{n+1}\right)^4b_{n+1}\geq 2b_{n+1}
$$
and by induction you have $a_n\geq 2b_n$ so $a_n-b_n\geq b_n\to\infty$ with $n\to\infty$.
A: The second term is very small compared to the first as $n$ gets large. Specifically
$$\lim_{n \rightarrow \infty} {4^n - n^4 \over 4^n} = \lim_{n \rightarrow \infty} (1 - {n^4 \over 4^n})$$
$$= 1 - \lim_{n \rightarrow \infty} {n^4 \over 4^n}$$
The limit on the right can be shown to be zero in several ways, such as applying L'hopital's rule several times. So the ratio goes to $1$ as $n$ goes to infinity. As a result,
$$\lim_{n \rightarrow \infty} 4^n - n^4 =  \lim_{n \rightarrow \infty} {4^n - n^4 \over 4^n} 4^n$$
$$= \lim_{n \rightarrow \infty} {4^n - n^4 \over 4^n}\lim_{n \rightarrow \infty} 4^n$$
$$= \infty$$
(The product rule for limits works for such limits).
Alternatively, the above limit shows that for $n$ large enough you have 
$${4^n - n^4 \over 4^n} > {1 \over 2}$$
This means
$$4^n - n^4 > {1 \over 2}4^n$$
So since $\lim _{n \rightarrow \infty} 4^n = \infty$, $\lim_{n \rightarrow \infty} 4^n - n^4 = \infty$ as well.
A: Consider the following form of your limit:
$$\lim_{n\rightarrow \infty} (4^{n}-n^{4})=\lim_{n\rightarrow \infty} (2^{n}-n^{2})(2^{n}+n^{2})\geq \lim_{n\rightarrow \infty}\left(2\dbinom{n}{1}+2\dbinom{n}{2} -n^{2}\right)(2^{n}+n^{2})=\lim_{n\rightarrow \infty}n(2^{n}+n^{2}) \longrightarrow \infty$$
Above i used the fact that: $$\sum_{k=0}^{n} \dbinom{n}{k}=2^n$$
Hence, the limit is $\infty$.
The proof is complete.
