<s>[INST] Incorrect solution in limits $$
\lim_{n\to \infty}\left(\frac{1}{n^4}{+3^{\frac{2}{2+n}}}\right)^{n}
$$
So i re-write it like:
$\lim_{n\to \infty}e^{n\ln{\frac{1}{n^4}\ln3^{\frac{2}{2+n}}}}$ $=$  $e^{\frac{2n}{2+n}\ln{\frac{1}{n^4}\ln3}}=e^{{2}\ln{\frac{1}{n^4}\ln3}}$
So here, $\ln{\frac{1}{n^4}}$ give  us minus infinity, but i think that somewhere i lost a way, so i need some hints
 [/INST] Put $n=\frac{1}{y}$ .The limit reduces to 
$L=\lim_{y\to 0}\left(y^4+3^{\frac{2y}{2y+1}}\right)^{\frac{1}{y}}$
Take log on both sides
$\log_{e}L=\lim_{y\to 0} \log_{e}\left(y^4+3^{\frac{2y}{2y+1}}\right)^{\frac{1}{y}}$
$\log_{e}L=\lim_{y\to 0}\frac{1}{y} \log_{e}\left(y^4+3^{\frac{2y}{2y+1}}\right)$
$\log_{e}L=\lim_{y\to 0}\frac{ \log_{e}\left(1+y^4+3^{\frac{2y}{2y+1}}-1\right)}{y^4+3^{\frac{2y}{2y+1}}-1}\frac{y^4+3^{\frac{2y}{2y+1}}-1}{y}$
$\log_{e}L=\lim_{y\to 0}\frac{ \log_{e}\left(1+y^4+3^{\frac{2y}{2y+1}}-1\right)}{y^4+3^{\frac{2y}{2y+1}}-1}\left(y^3+\frac{3^{\frac{2y}{2y+1}}-1}{\frac{2y}{2y+1}}\frac{2}{2y+1}\right)=1\left(2\log_{e}3\right)=\log_{e}9$
As $y\to 0$ it is obvious that $y^4+3^{\frac{2y}{2y+1}}-1\to 0$. 
So here  the standard result $\lim_{x\to 0}\frac{\log_{e}(1+x)}{x}=1$ has been used. 
Also as $y\to 0$ it is obvious that $\frac{2y}{2y+1}\to 0$ 
so the standard limit $\lim_{x\to 0}\frac{a^x-1}{x}=\log_{e}a$ can be used.
Hence,$L=9$
</s>