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 inequalites are wrong. To make a fraction larger you have to make the numerator larger  and the denominator smaller. To make the fraction smaller make the numerator smaller and the denominator larger.
To compute the limit, observe that the dominating term in the numerator and denominator is $4^n/n^4$. The other terms are neglegible.
