<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] If you want more than the limit itself
$$y_n= \left(1+ \frac{3}{n^2+n^4}\right)^n\implies \log(y_n)=n \log\left(1+ \frac{3}{n^2+n^4}\right)$$
Using Taylor expansion
$$ \log\left(1+ \frac{3}{n^2+n^4}\right)=\frac{3}{n^4}-\frac{3}{n^6}+O\left(\frac{1}{n^8}\right)$$
$$\log(y_n)=\frac{3}{n^3}-\frac{3}{n^5}+O\left(\frac{1}{n^7}\right)$$
$$y_n=e^{\log(y_n)}=1+\frac{3}{n^3}-\frac{3}{n^5}+\frac{9}{2n^6}+O\left(\frac{1}{n^7}\right)$$ which shows the limit and how it is approached.
Moreover, this gives you a shortcut for the evaluation of $y_n$.
Consider $n=5$ and use your pocket calculator
$$y_5=\frac{118731486838493}{116029062500000}\approx 1.023291$$ while the above truncated expansion gives $\frac{31979}{31250}=1.023328$ (corresponding to a relative error equal to $3.6 \times 10^{-3}$%).
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