$\sqrt[n]{\vert \frac{1}{n^{10}3^{n}} - \frac{n^{10}}{e^n} \vert}, n \in N$
The only thing I figure out is:
$\sqrt[n]{\vert \frac{e^n - n^{100}3^n}{n^{10}3^{n}e^{n}} \vert} = \frac{\sqrt[n]{\vert {e^n - n^{100}3^n} \vert}}{\sqrt[n]{n^{10}3^{n}e^{n}}} = \frac{\sqrt[n]{\vert {e^n - n^{100}3^n} \vert}}{3e\sqrt[n]{n^{10}}} $
I've made a mistake in nominator, now it's correct.
if we continue from the algebraic manipulation you've offered, it is clear that $\lim_{n\rightarrow\infty}n^{1/n}=1$ and therefore the denominator approaches $3e$ as $n$ goes to $\infty$. for the enumarator, it is clear that for $n\geq2$ , $3^nn^{100}/2>e^n$, so for $n\geq2$ we get $\sqrt[n]{\frac{1}{2}3^{n}n^{100}}\le\sqrt[n]{3^{n}n^{100}-e^{n}}\le\sqrt[n]{\frac{3}{2}3^{n}n^{100}}$ and since for all $c\in\mathbb{R}$, $\lim_{n\rightarrow\infty}\sqrt[n]{c}=1$, and $\lim_{n\rightarrow\infty}\sqrt[n]{n}=1$ squeeze theorem gives gives that the enumerator approches 3, and in conclusion the limit is $3/3e=1/e$
