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 in your approach is this.
$a,b,c,d>0$, if $a>b$ and $c>d$ you cannot conclude $\frac{a}{c}>\frac{b}{d}$
