<s>[INST] Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule 
Find the following limit
  $$\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$$
  without using L'Hopital's rule.

I tried to solve this using fundamental limits such as $\lim_{x\to0}\left(1+x\right)^\frac1x=e,\lim_{x\to0}x^x=1$ and equivalent infinitesimals at $x\to0$ such as $x\sim\sin x,a^x\sim1+x\ln a,(1+x)^a\sim1+ax$. This is what I did so far:
$$\begin{align}\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\lim_{x\to0}\frac{\left(\frac{3-(1-\cos x)}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(\frac{3-\frac12x^2}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(1-\frac16x^2\right)^x-1}{x^3}\end{align}$$
I used fact that $x\sim\sin x$ at $x\to0$. After that, I tried to simplify $\left(1-\frac16x^2\right)^x$. Problem is because this is not indeterminate, so I cannot use infinitesimals here. Can this be solved algebraically using fundamental limits, or I need different approach?
 [/INST] We can proceed in the following manner
\begin{align}
L &= \lim_{x \to 0}\frac{1}{x^{3}}\left(\left(\frac{2 + \cos x}{3}\right)^{x} - 1\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{3}}\left(\exp\left(x\log\left(\frac{2 + \cos x}{3}\right)\right) - 1\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{3}}\cdot x\log\left(\dfrac{2 + \cos x}{3}\right)\cdot\dfrac{\exp\left(x\log\left(\dfrac{2 + \cos x}{3}\right)\right) - 1}{x\log\left(\dfrac{2 + \cos x}{3}\right)}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \log\left(\dfrac{2 + \cos x}{3}\right)\cdot 1\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \log\left(1 + \frac{\cos x - 1}{3}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot \dfrac{\cos x - 1}{3}\cdot\dfrac{\log\left(1 + \dfrac{\cos x - 1}{3}\right)}{\dfrac{\cos x - 1}{3}}\notag\\
&= -\frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\notag\\
&= -\frac{1}{6}\notag
\end{align}
We have used the following standard limits $$\lim_{x \to 0}\frac{\exp(x) - 1}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}$$
</s>