Evaluating the limit $\lim_{x \to 0}\left(x+e^{\frac{x}{3}}\right){}^{\!\frac{3}{x}}$ 
$$y=\left(x+e^{\frac{x}{3}}\right)^{\frac{3}{x}}$$

I'm looking at this equation, and need to solve for the limit as $ \to 0$. I've graphed it, and know the solution is $e^4$, but am clueless as to the steps to actually solve this. 
(Note, I am an adult working as a math aide in a high school. I help students at Algebra, Trig, Geometry, and Intro to Calculus. 35 years out of HS myself, there are clearly some things I need to brush up on. i.e. I know my limits. Pun intended.)
Yes, L'Hopital is fine. The student is in a BC calc class)

If $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(x + e^{x/3}\right)^{3/x}\right\}\notag\\
&= \lim_{x \to 0}\log\left(x + e^{x/3}\right)^{3/x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{3}{x}\cdot\log\left(x + e^{x/3}\right)\notag\\
&= \lim_{x \to 0}\frac{3}{x}\cdot\log\left(1 + x + e^{x/3} - 1\right)\notag\\
&= \lim_{x \to 0}\frac{3}{x}\cdot (x + e^{x/3} - 1)\cdot\frac{\log\left(1 + x + e^{x/3} - 1\right)}{x + e^{x/3} - 1}\notag\\
&= \lim_{x \to 0}\frac{3}{x}\cdot (x + e^{x/3} - 1)\cdot\lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = x + e^{x/3} - 1)\notag\\
&= \lim_{x \to 0}\left(3 + \frac{e^{x/3} - 1}{x/3}\right)\notag\\
&= 3 + \lim_{t \to 0}\frac{e^{t} - 1}{t}\text{ (putting }t = x/3)\notag\\
&= 3 + 1 = 4\notag
\end{align}
and hence $L = e^{4}$.
