$ \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?
 A: Yes, I think it's right. Of course, you might have to justify that your approximations still preserve the limit. 
Another way would be to write 
$$
\left(1+ \frac{3}{n^2+n^4}\right)^n=\exp\left( n\log\left(1+ \frac{3}{n^2+n^4}\right)\right)\sim\exp\left( \frac{3n}{n^2+n^4}\right)\to e^0=1.
$$
A: 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}
A: 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,
$$1\leq \left(1 +\frac{3}{n^2+n^4}\right)^n\leq \left(1 +\frac{3}{n^4}\right)^n$$
$$\leq 1+\sum_{k=1}^n\left( \frac{3}{n^4}\right)^k\binom nk \stackrel{(\star)}{\leq} \sum_{k=1}^n\left( \frac{3}{n^3}\right)^k\frac 1{k!}$$
$$\leq 1 + \frac 1{n^3}\sum_{k=1}^n\frac{3^k}{k!} \leq 1+\frac{e^3-1}{n^3}\stackrel{n\to\infty}{\longrightarrow}1$$
A: 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 $3.6 \times 10^{-3}$%).
