$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$     \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$    (1+ \frac{3}{n^2+n^4})^n =   \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim    \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim   \sqrt [n^3]{e^3 }  \rightarrow 1$
Is it right?

It is more or less correct, but how do you know that$$\lim_{n\to\infty}\left(1+\frac3{n^2+n^4}\right)^{n^4}=e^3?$$
It seems to me that it is more natural to do it as follows:\begin{align}\lim_{n\to\infty}\left(1+\frac3{n^2+n^4}\right)^{n^3}&=\lim_{n\to\infty}\left(\left(1+\frac3{n^2+n^4}\right)^{n^2+n^4}\right)^{\frac{n^3}{n^2+n^4}}\\&=(e^3)^0\\&=1.\end{align}
