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)

Hint Provided that $\lim_{x \to 0} \log y$ exists, by continuity we have
$$e^{\lim_{x \to 0} \log y} = \lim_{x \to 0} e^{\log y} = \lim_{x \to 0} y,$$ and on the other hand,
$$\log y = 3 \cdot \frac{\log(x + e^{x / 3})}{x}.$$
Now, we can evaluate the limit of $\log y$ as $x \to 0$ by (1) applying L'Hopital's Rule, or (2) using some elementary Taylor series expansions to conclude that that $$x + e^{x / 3} = 1 + \frac{4}{3} x + O(x^2)$$
and hence
$$\log(x + e^{x / 3}) = \frac{4}{3} x + O(x^2) .$$
