Limit of $\frac{2^n+2.71^n+\frac{4^n}{n^4}}{\frac{4^n}{n^4}+n^33^n}$ - what is wrong with my proof? Here's quite a hairy sequence, the limit of which I need to find as $n\rightarrow\infty$:
$$\frac{2^n+2.71^n+\frac{4^n}{n^4}}{\frac{4^n}{n^4}+n^33^n}$$
The Squeeze Theorem seemed like a good idea so here's what I've done:
$\frac{2^n+2^n+2^n}{2^n+2^n}<=\frac{2^nn^4+2.71^nn^4+4^n}{4^n+n^73^n}=\frac{2^n+2.71^n+\frac{4^n}{n^4}}{\frac{4^n}{n^4}+n^33^n}<=\frac{n^n+n^n+n^n}{n^n+n^n}$
The left and right sequences have the $\lim_{n\rightarrow\infty}=\frac{3}{2}$. So theoretically the sequence for which $\lim$ is being searched should also converge to $\frac{3}{2}$. But WolframAlpha tells me that the correct answer is $1$. So where is an error in my logic?
One more question, in my "proof" I couldn't show that for sufficiently large $n$: $n^33^n<=n^n$ and this is important in order for my transformation to be considered as correct. Is there any way to show it without derivatives?

The problem with your inequalities is you have bounded both the numerator and denominator below and then both above. I.e., It's not the case that $a/b < c/d$ if $0 < a < c$ and $0 < b < d$.
Hint: Multiple top and bottom by $n^4.4^{-n}$. You have a $1$ in both places and all other terms can be shown to tend to zero.
