lim calculus problem with infinity $$\lim_{n→∞} \left( \left(\frac{n+3}{n+5}\right)^{n+4} + \sqrt[2n]{3n}\right)$$

I tried replacing n+5 with u and simplifying but the answer seems off. Help would be greatly appreciated.
We need to do some rewriting:

$$\begin{align}
\left(\frac{n+3}{n+5}\right)^{n+4} &amp;= \left(\frac{n+5-2}{n+5}\right)^{n+4} \\
&amp;= \left(1-\frac{2}{n+5}\right)^{n+5-1} \\
&amp;= \underbrace{ \left(1-\frac{2}{n+5}\right)^{n+5} }_{\to e^{-2}}\ \underbrace{ \left(1-\frac{2} {n+5}\right)^{-1} }_{\to 1}
\end{align}
$$

So the left term goes to $e^{-2}$. The right side can be written as $(\sqrt[n]{3n})^{\frac12}$. The $n$-th root of any polynomial in $n$ goes to $1$, so this goes to $1$. The overall limit is $e^{-2}-1$.
The inner fraction simplifies to 1, so 1 to infinity is 1. 1 + infinity = infinity.

(then again it is quite difficult to understand from the formatting)

P.S. After replacing u = n+5, you get $\frac{u - 2}{u}$, which isn't off.