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)
 A: $$y=\left(x+e^{\frac{x}{3}}\right)^{\frac{3}{x}}$$ $$\log(y)={\frac{3}{x}}\log\left(x+e^{\frac{x}{3}}\right)$$ Now, using that, close to $t=0$, $e^t\simeq 1+t$.
So $$\log(y)\simeq {\frac{3}{x}}\log\left(x+1+{\frac{x}{3}}\right)={\frac{3}{x}}\log\left(1+{\frac{4x}{3}}\right)$$ Now, using that, close to $t=0$ $\log(1+t)\sim t$, $$\log(y)\simeq {\frac{3}{x}}{\frac{4x}{3}}=4$$
A: Intuition:
A way to see what is going on is to see the affine approximation of $e^x$ around $0$: $$e^u \simeq e^0 + (e^\prime)(0) x  = 1 + x$$ (this can be made formal by Taylor approximations to order $1$, for instance). This implies that your quantity is roughly $\left(x+ 1+ \frac{x}{3}\right)^{3/x} = \left(1+ \frac{4x}{3}\right)^{3/x}$, where you recognize, setting $t = \frac{3}{x}\to \infty$, the limit $$\left(1+\frac{4}{t}\right)^t \xrightarrow[t\to\infty]{} e^4.$$ The only key is to make this first approximation $\simeq$ rigorous, which is done below.
An approach based on Taylor expansions: (but which requires no knowledge of them besides the Landau notation $o(\cdot)$ — justifying what is needed as we go)
Start (as often when you have both a base and an exponent depending on $x$) by rewriting it in exponential form:
$$
\left(x+e^{\frac{x}{3}}\right)^\frac{3}{x} = e^{\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right)} 
$$
Now, when $u\to 0$, we have $\frac{e^u-1}{u}\to \exp^\prime 0 = e^0 = 1$, so that $e^u = 1+u + o(u)$; which gives $$x+e^{\frac{x}{3}} = x+1+ \frac{x}{3} + o(x) = 1+\frac{4}{3}x.$$
Similarly, since $\frac{\ln(1+u)}{u}\xrightarrow[u\to 0]{} 1$, we have $\ln(1+u) =  u + o(u)$. Combining the two, we get 
$$\ln\left(x+e^{\frac{x}{3}}\right) = \ln\left(1+\frac{4}{3}x\right) = \frac{4}{3}x + o(x).$$
Putting it together, 
$$
\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right) = \frac{3}{x}\left(\frac{4}{3}x + o(x)\right) = 4 + o(1) \xrightarrow[x\to 0]{} 4
$$
and, by continuity of $\exp$,
$$e^{\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right)} \xrightarrow[x\to 0]{} e^4.
$$
A: 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
$$\log(x + e^{x / 3}) = \frac{4}{3} x + O(x^2) .$$
A: 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 $L = e^{4}$.
