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
 A: 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,$L=9$
A: Note that
$$
\mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{a}
{n}} \right)^{\,n}  = e^{\,a} \quad \quad \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{f(n)}}
{n}} \right)^{\,n}  \ne \mathop {\lim }\limits_{n\; \to \;\infty } e^{\,f(n)} 
$$
that's the fault in your derivation. Consider instead that we have:
$$
\mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{a}
{{\left( {n + \alpha } \right)}} + \frac{b}
{{\left( {n + \beta } \right)^{\,2} }} +  \cdots } \right)^{\,n + \gamma }  = e^{\,a} 
$$
Therefore we shall try and expand $3^{2/(2+n)}$ in powers of the exponent, i.e. Taylor of $3^x \quad | \, x=2/(2+n) \approx 0$
$$
\begin{gathered}
  \mathop {\lim }\limits_{n\; \to \;\infty } \left( {\frac{1}
{{n^{\,4} }} + 3^{\,\frac{2}
{{2 + n}}} } \right)^{\,n}  = \mathop {\lim }\limits_{n\; \to \;\infty } \left( {e^{\frac{{2\ln 3}}
{{2 + n}}}  + \frac{1}
{{n^{\,4} }}} \right)^{\,n}  = \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{2\ln 3}}
{{2 + n}} + \frac{{\left( {2\ln 3} \right)^2 }}
{2}\frac{1}
{{\left( {2 + n} \right)^2 }} +  \cdots  + \frac{1}
{{n^{\,4} }}} \right)^{\,n}  =  \hfill \\
   = \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{2\ln 3}}
{{2 + n}}} \right)^{\,n}  = e^{2\ln 3}  = 9 \hfill \\ 
\end{gathered} 
$$
A: We can see that it is " tending to one" raised to the power "tending to infinity" then we can use this as a shortcut
Then use L'Hospital rule to simplify
