<s>[INST] $ \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?
 [/INST] You may also just use the binomial formula and squeeze by applying 


*

*$(\star)$: $\frac 1{n^k}\binom nk \leq \frac 1{k!}$ for $0\leq k\leq n$
Hence,
$$1\leq \left(1 +\frac{3}{n^2+n^4}\right)^n\leq \left(1 +\frac{3}{n^4}\right)^n$$
$$\leq 1+\sum_{k=1}^n\left( \frac{3}{n^4}\right)^k\binom nk \stackrel{(\star)}{\leq} \sum_{k=1}^n\left( \frac{3}{n^3}\right)^k\frac 1{k!}$$
$$\leq 1 + \frac 1{n^3}\sum_{k=1}^n\frac{3^k}{k!} \leq 1+\frac{e^3-1}{n^3}\stackrel{n\to\infty}{\longrightarrow}1$$
</s>